CE Board Exam Randomizer

Question 1

$$\frac{\partial F}{\partial x}\frac{dx}{dt} + \frac{\partial F}{\partial y}\frac{dy}{dt} + \frac{\partial F}{\partial t} = 0$$

Answer

Let $x$ = distance of foot from wall, $y$ = height of top on wall. By Pythagorean theorem:

$x^2 + y^2 = 100$

When $x = 6$: $y = \sqrt{100 - 36} = 8$ m.

Differentiate with respect to time:

$2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0$

$\dfrac{dy}{dt} = -\dfrac{x}{y}\cdot\dfrac{dx}{dt} = -\dfrac{6}{8}(2) = -\dfrac{3}{2}$

The top slides down at $\boxed{1.5\ \text{m/s}}$. (Negative sign indicates downward.)

Answer

By similar triangles: $\dfrac{r}{h} = \dfrac{4}{8} = \dfrac{1}{2}$, so $r = \dfrac{h}{2}$.

Volume of a cone:

$V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi\!\left(\dfrac{h}{2}\right)^2\!h = \dfrac{\pi h^3}{12}$

Differentiate with respect to time:

$\dfrac{dV}{dt} = \dfrac{\pi h^2}{4}\cdot\dfrac{dh}{dt}$

Solve for $\dfrac{dh}{dt}$ (with $dV/dt = -2$, $h = 4$):

$\dfrac{dh}{dt} = \dfrac{4}{\pi h^2}\cdot\dfrac{dV}{dt} = \dfrac{4}{\pi(16)}(-2) = -\dfrac{1}{2\pi}$

The water level drops at $\boxed{\dfrac{1}{2\pi} \approx 0.159\ \text{m/min}}$.

Answer

$A = \pi r^2$. Differentiate with respect to time:

$\dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt}$

Substitute $r = 5$ cm and $\dfrac{dr}{dt} = 3$ cm/s:

$\dfrac{dA}{dt} = 2\pi(5)(3) = \boxed{30\pi \approx 94.25\ \text{cm}^2/\text{s}}$

Answer

After 1 hour: $y = 60$ km (north), $x = 80$ km (east).

$d = \sqrt{x^2 + y^2} = \sqrt{80^2 + 60^2} = \sqrt{10000} = 100\ \text{km}$

Differentiate $d^2 = x^2 + y^2$:

$2d\dfrac{dd}{dt} = 2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt}$

$\dfrac{dd}{dt} = \dfrac{x\,\dot{x} + y\,\dot{y}}{d} = \dfrac{80(80) + 60(60)}{100} = \dfrac{6400 + 3600}{100} = \boxed{100\ \text{km/h}}$

Answer

Let $x$ = distance from lamp to person, $s$ = length of shadow. By similar triangles:

$\dfrac{10}{x + s} = \dfrac{2}{s} \implies 10s = 2(x+s) \implies s = \dfrac{x}{4}$

The tip of the shadow is at distance $T = x + s = x + \dfrac{x}{4} = \dfrac{5x}{4}$ from the lamp.

$\dfrac{dT}{dt} = \dfrac{5}{4}\dfrac{dx}{dt} = \dfrac{5}{4}(1.5) = \boxed{1.875\ \text{m/s}}$

The shadow length itself grows at $ds/dt = \dfrac{1}{4}(1.5) = 0.375$ m/s.

Answer

For a sphere:
$V=\frac{4}{3}\pi r^3$, so $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$
At $r=4$ and $dV/dt=60\pi$:
$60\pi=4\pi(4^2)\frac{dr}{dt}$, so $dr/dt=15/16$.
Surface area $S=4\pi r^2$:
$\frac{dS}{dt}=8\pi r\frac{dr}{dt}=8\pi(4)(15/16)$
$\boxed{30\pi}$

Answer

Let the variable leg be $x$ and the hypotenuse be $h$. Then:
$h^2=x^2+6^2$
Differentiate:
$2h\frac{dh}{dt}=2x\frac{dx}{dt}$
When $h=10$, $x=8$. With $dx/dt=2$:
$\frac{dh}{dt}=\frac{8(2)}{10}=1.6$
$\boxed{1.6\text{ ft/s}}$

Answer

At water depth $h$, the surface width is $2+h$. Cross-sectional area is:
$A=\frac{h[2+(2+h)]}{2}=2h+\frac{h^2}{2}$
Volume $V=10A=20h+5h^2$. Differentiate:
$\frac{dV}{dt}=(20+10h)\frac{dh}{dt}$
At $h=6$ in $=0.5$ ft and $dV/dt=3$:
$3=25\frac{dh}{dt}$
$\boxed{0.12\text{ ft/min}}$

Answer

For similar triangles, $r/h=5/12$. Thus:
$V=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi\left(\frac{5h}{12}\right)^2h$
Differentiate:
$\frac{dV}{dt}=\pi\frac{25}{144}h^2\frac{dh}{dt}$
At $h=6$ and $dV/dt=10$:
$10=\pi\frac{25}{144}(36)\frac{dh}{dt}$
$\boxed{0.51\text{ m/s}}$

Answer

Let $x$ be the man's distance from the point below the light and $s$ be the shadow length. Similar triangles give:
$\frac{15}{x+s}=\frac{6}{s}$
$15s=6x+6s$, so $9s=6x$ and $s=\frac{2}{3}x$. The shadow tip position is $x+s=\frac{5}{3}x$, so:
$\frac{d}{dt}(x+s)=\frac{5}{3}\frac{dx}{dt}=\frac{5}{3}(3)$
$\boxed{5\text{ mph}}$

Answer

Surface area after 2 seconds is:
$A=26(2)=52$ cm². Since $A=4\pi r^2$:
$r=\sqrt{52/(4\pi)}=\sqrt{13/\pi}$
Differentiate $A=4\pi r^2$:
$26=8\pi r\frac{dr}{dt}$
$\boxed{\frac{dr}{dt}=0.509\text{ cm/s}}$

Answer

For the cone, $r/h=3.75/15=1/4$, so $r=h/4$. Volume:
$V=\frac{1}{3}\pi r^2h=\frac{\pi h^3}{48}$
Differentiate:
$\frac{dV}{dt}=\frac{\pi h^2}{16}\frac{dh}{dt}$
At $h=4$ ft and $dh/dt=2$ ft/min:
$\frac{dV}{dt}=\frac{\pi(16)}{16}(2)$
$\boxed{6.28}$

Answer

Let $x$ be the lower-end distance from the wall and $y$ the upper-end height. The pipe length is 3 m = 300 cm:
$x^2+y^2=300^2$
When $x=2$ m = 200 cm, $y=\sqrt{300^2-200^2}=223.61$ cm. Differentiate:
$x\frac{dx}{dt}+y\frac{dy}{dt}=0$
$\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}=-\frac{200}{223.61}(2)$
$\boxed{1.79\text{ cm/s downward}}$

Answer

For the cone, $r/h=3.75/15=1/4$, so $r=h/4$. Volume:
$V=\frac{1}{3}\pi r^2h=\frac{\pi h^3}{48}$
$\frac{dV}{dt}=\frac{\pi h^2}{16}\frac{dh}{dt}$
At $h=4$ ft and $dh/dt=2$ ft/min:
$\frac{dV}{dt}=\frac{\pi(4)^2}{16}(2)=2\pi$
$\boxed{6.28}$

Answer

Surface area of the pool is:
$A=\pi r^2=\pi(3)^2=9\pi$ m²
Evaporation rate:
$2(9\pi)=18\pi$ L/hr
Time for 30 L:
$t=30/(18\pi)=0.5305$ hr
$0.5305(60)=31.8$ min
$\boxed{32\text{ min}}$

Answer

Part 1.

Height $h = 12 + 10\sin\varphi$. At $h = 17$: $\sin\varphi = 0.5 \Rightarrow \varphi = 30^\circ$.
Vertical speed $\frac{dh}{dt} = 10\cos\varphi\frac{d\varphi}{dt} = 2.52$:
$\frac{d\varphi}{dt} = \frac{2.52}{10\cos30^\circ} = 0.291$ rad/s
Convert to rev/min: $0.291\times\frac{60}{2\pi}$
$\boxed{2.78 \text{ rpm}}$

Part 2.

Horizontal position $x = 10\cos\varphi$, so $\frac{dx}{dt} = -10\sin\varphi\frac{d\varphi}{dt}$.
At $\varphi = 30^\circ$ with $\frac{d\varphi}{dt} = 0.291$ rad/s:
$\frac{dx}{dt} = -10(0.5)(0.291)$
$\boxed{1.46 \text{ m/s}}$

Part 3.

At the top, the centripetal acceleration points downward: $a_c = \omega^2 r = (0.291)^2(10) = 0.847$ m/s².
The seat force (normal) is $N = mg - ma_c$:
$N = 75(9.81) - 75(0.847)$
$\boxed{672 \text{ N}}$

Answer

Let $x = 100 - 50t$ and $y = 150 - 50t$ be the trains' distances from the crossing. At $t = 3.5$ h: $x = -75$, $y = -25$.
$s = \sqrt{75^2 + 25^2} = 79.06$ km
$s\frac{ds}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt} = (-75)(-50) + (-25)(-50) = 5000$
$\frac{ds}{dt} = \frac{5000}{79.06}$
$\boxed{63.25 \text{ kph}}$

Answer

The rope is the hypotenuse $L$, with vertical leg $4.2$ m and horizontal leg $x$.
At $L = 6.5$: $x = \sqrt{6.5^2 - 4.2^2} = 4.961$ m.
From $L^2 = 4.2^2 + x^2$: $L\frac{dL}{dt} = x\frac{dx}{dt}$.
With $\frac{dL}{dt} = -0.6$ m/s:
$\frac{dx}{dt} = \frac{6.5(-0.6)}{4.961}$
$\boxed{0.786 \text{ m/s}}$

Answer

Part 1.

Integrate the accelerations (from rest): $v_x = 0.4t^2$, $v_y = 2t - 0.15t^2$, $v_z = 5t$.
At $t = 10$: $v_x = 40$, $v_y = 5$, $v_z = 50$.
$|v| = \sqrt{40^2 + 5^2 + 50^2} = \sqrt{4125}$
$\boxed{64.23 \text{ m/s}}$

Part 2.

Total distance is $\int_0^{10}|v|\,dt = \int_0^{10}\sqrt{(0.4t^2)^2 + (2t - 0.15t^2)^2 + (5t)^2}\,dt$.
Evaluating numerically (e.g. Simpson's rule):
$\boxed{\approx 292 \text{ m}}$

Part 3.

At $t = 10$: $a_x = 0.8(10) = 8$, $a_y = 2 - 0.3(10) = -1$, $a_z = 5$.
$|a| = \sqrt{8^2 + (-1)^2 + 5^2} = \sqrt{90}$
$\boxed{9.49 \text{ m/s}^2}$

Part 4.

Integrate the velocities to get position (from origin): $x = \frac{0.4t^3}{3}$, $y = t^2 - 0.05t^3$, $z = 2.5t^2$.
At $t = 10$: $x = 133.3$, $y = 50$, $z = 250$.
Displacement $= \sqrt{133.3^2 + 50^2 + 250^2}$
$\boxed{288 \text{ m}}$

Answer

Each section is a circle of diameter $d = 20y - 3y^2$, so $dV = \pi\left(\frac{d}{2}\right)^2 dy = \frac{\pi}{4}(20y - 3y^2)^2\,dy$.
$V = \frac{\pi}{4}\int_0^4 (400y^2 - 120y^3 + 9y^4)\,dy$
$= \frac{\pi}{4}\left[\frac{400y^3}{3} - 30y^4 + \frac{9y^5}{5}\right]_0^4 = \frac{\pi}{4}(2696.5)$
$\boxed{2117.85}$

Answer

With apex down, the radius at height $z$ is $\frac{z}{4}$, so a slice has volume $\pi\frac{z^2}{16}\,dz$ and is lifted $(2 - z)$ to the top.
$W = \rho g\pi\int_0^2 \frac{z^2}{16}(2 - z)\,dz = \frac{9810\pi}{16}\left[\frac{2z^3}{3} - \frac{z^4}{4}\right]_0^2$
$= \frac{9810\pi}{16}(1.333)$
$\boxed{2.568 \text{ kN-m}}$

Answer

The source statement is missing the rate at which the juice level in the funnel is falling. The keyed answer corresponds to the funnel level falling at 1.25 cm/s.
For the cone, $r/h=6/20=0.3$, so
$$V_f=\frac{1}{3}\pi(0.3h)^2h=0.03\pi h^3$$
Differentiate:
$$\frac{dV_f}{dt}=0.09\pi h^2\frac{dh}{dt}$$
At $h=10$ cm:
$$\frac{dV_f}{dt}=9\pi\frac{dh}{dt}$$
The cylinder radius is 5 cm, so
$$\frac{dV_c}{dt}=25\pi\frac{dH}{dt}$$
Since volume leaving the funnel enters the cylinder,
$$25\pi\frac{dH}{dt}=-9\pi\frac{dh}{dt}$$
Using $dh/dt=-1.25$ cm/s:
$$\frac{dH}{dt}=0.45$$
$$\boxed{0.45}$$

Answer

In 30 seconds, the plane travels
$$30\text{ sec}=\frac{1}{120}\text{ hr}$$
$$x=240\left(\frac{1}{120}\right)=2\text{ km}$$
Let s be the distance from the observer. With altitude 1 km,
$$s^2=x^2+1^2$$
Differentiate:
$$2s\frac{ds}{dt}=2x\frac{dx}{dt}$$
At $x=2$ km, $s=\sqrt{5}$ km and $dx/dt=240$ kph:
$$\frac{ds}{dt}=\frac{2}{\sqrt{5}}(240)=214.66$$
$$\boxed{214.66\text{ kph}}$$

Answer

Convert the radius rate:
$$20\text{ mm/sec}=0.020\text{ m/sec}$$
The area of one circular face is
$$A=\pi r^2$$
Differentiate:
$$\frac{dA}{dt}=2\pi r\frac{dr}{dt}$$
At $r=8.1$ m:
$$\frac{dA}{dt}=2\pi(8.1)(0.020)=1.018$$
$$\boxed{1.018\text{ sq m/sec}}$$

Answer

Let y be the elevator height above the horizontal line of sight, and let $\theta$ be the angle of elevation. The spectator is 20 m from the shaft.
$$\tan\theta=\frac{y}{20}$$
Differentiate:
$$\sec^2\theta\frac{d\theta}{dt}=\frac{1}{20}\frac{dy}{dt}$$
Six seconds after the line of sight was horizontal:
$$y=5(6)=30\text{ m}$$
So
$$\tan\theta=\frac{30}{20}=\frac{3}{2}$$
$$\sec^2\theta=1+\tan^2\theta=1+\frac{9}{4}=\frac{13}{4}$$
Then
$$\frac{13}{4}\frac{d\theta}{dt}=\frac{5}{20}=\frac{1}{4}$$
$$\frac{d\theta}{dt}=\boxed{\frac{1}{13}}$$

Answer

Let x be the boy's distance past the crossing point on the bridge, y be the automobile's distance past the crossing point on the highway, and 24 m be the vertical separation.
$$s^2=x^2+y^2+24^2$$
Differentiate:
$$2s\frac{ds}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$$
At the instant described, $x=6$, $y=8$, $dx/dt=6$, and $dy/dt=80$.
$$s=\sqrt{6^2+8^2+24^2}=26$$
Thus,
$$\frac{ds}{dt}=\frac{6(6)+8(80)}{26}=26$$
$$\boxed{26\text{ m/s}}$$

Answer

The printed equation appears incomplete; the keyed answer matches the parabola $y^2=8x$.
Differentiate with respect to time:
$$2y\frac{dy}{dt}=8\frac{dx}{dt}$$
Given $dy/dt=5$ units/sec and $y=4$:
$$2(4)(5)=8\frac{dx}{dt}$$
$$\frac{dx}{dt}=5$$
$$\boxed{5}$$

Answer

The time available is the time until either plane reaches the convergence point.
For the first plane:
$$t=\frac{150}{450}=\frac{1}{3}\text{ hr}$$
For the second plane:
$$t=\frac{200}{600}=\frac{1}{3}\text{ hr}$$
Convert to minutes:
$$\frac{1}{3}\text{ hr}=20\text{ min}$$
$$\boxed{20\text{ min}}$$

Answer

Let x be the LRT's horizontal distance from the crossing, y be the car's distance from the crossing, and 6 m be the vertical separation.
One second later:
$$x=9(1)=9$$
$$y=12-4(1)=8$$
The distance s between them satisfies
$$s^2=x^2+y^2+6^2$$
Differentiate:
$$2s\frac{ds}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$$
Here $dx/dt=9$ and $dy/dt=-4$. Also,
$$s=\sqrt{9^2+8^2+6^2}=\sqrt{181}$$
Thus,
$$\frac{ds}{dt}=\frac{9(9)+8(-4)}{\sqrt{181}}=3.64$$
$$\boxed{3.64\text{ m/s}}$$

Answer

The source statement is garbled and does not give a fully reliable diagram or light height. The problem is a similar-triangles shadow-rate setup: define the shadow height on the wall as y, relate y to the man's distance x from the wall using the light-wall geometry, then differentiate $y=f(x)$ and substitute $dx/dt=-0.6$ m/s at $x=3$ m.
The source answer key gives the shortening rate as
$$\boxed{-0.192\text{ m/s}}$$

Answer

The light is 12 ft high, and the vendor walks along the curb 12 ft horizontally from the light. Let x be the vendor's distance along the curb. The vendor's speed is
$$420\text{ fpm}=7\text{ fps}$$
Using similar triangles in 3D, the ray from the 12-ft light through the vendor's 5-ft head reaches the ground at a scale factor
$$\lambda=\frac{12}{12-5}=\frac{12}{7}$$
Thus the shadow tip's coordinate along the curb direction is
$$X=\frac{12}{7}x$$
Differentiate:
$$\frac{dX}{dt}=\frac{12}{7}\frac{dx}{dt}=\frac{12}{7}(7)=12$$
$$\boxed{12\text{ fps}}$$

Answer

For an equilateral triangle with side s,
$$A=\frac{\sqrt{3}}{4}s^2$$
Differentiate:
$$\frac{dA}{dt}=\frac{\sqrt{3}}{2}s\frac{ds}{dt}$$
Given $dA/dt=100$ sq m/sec and $ds/dt=10$ m/s:
$$100=\frac{\sqrt{3}}{2}s(10)$$
$$100=5\sqrt{3}s$$
$$s=\frac{20}{\sqrt{3}}$$
$$\boxed{\frac{20}{\sqrt{3}}}$$

Answer

Using the cm-based dimensions, the top radius-to-height ratio is
$$\frac{r}{h}=\frac{3.75}{15}=\frac{1}{4}$$
$$r=\frac{h}{4}$$
The volume is
$$V=\frac{1}{3}\pi r^2h=\frac{\pi h^3}{48}$$
Differentiate:
$$\frac{dV}{dt}=\frac{\pi h^2}{16}\frac{dh}{dt}$$
At $h=4$ cm and $dh/dt=2$ cm/s:
$$\frac{dV}{dt}=\frac{\pi(4)^2}{16}(2)=2\pi=6.28$$
The source choices label the unit incorrectly, but the keyed numeric answer is
$$\boxed{6.28}$$

Answer

Let c be the third side. By the law of cosines,
$$c^2=5^2+8^2-2(5)(8)\cos\theta$$
$$c^2=89-80\cos\theta$$
At $\theta=60^\circ$:
$$c=\sqrt{89-80(0.5)}=7$$
Differentiate:
$$2c\frac{dc}{dt}=80\sin\theta\frac{d\theta}{dt}$$
With $d\theta/dt=1$ rad/sec:
$$\frac{dc}{dt}=\frac{40\sin60^\circ}{7}=4.95$$
$$\boxed{4.95\text{ units/sec}}$$

Answer

The area of a triangle with two sides and included angle is
$$A=\frac{1}{2}ab\sin\theta$$
With sides 5 m and 8 m:
$$A=20\sin\theta$$
Differentiate:
$$\frac{dA}{dt}=20\cos\theta\frac{d\theta}{dt}$$
At $\theta=60^\circ$ and $d\theta/dt=2$ rad/sec:
$$\frac{dA}{dt}=20\cos60^\circ(2)=20$$
$$\boxed{20\text{ sq m/sec}}$$

Answer

For the triangular end, width is proportional to depth. Since the trough is 2 m wide at 2 m deep,
$$w=h$$
The cross-sectional water area is
$$A=\frac{1}{2}wh=\frac{1}{2}h^2$$
The trough is 12 m long, so
$$V=12A=6h^2$$
Differentiate:
$$\frac{dV}{dt}=12h\frac{dh}{dt}$$
At $h=1$ m and $dV/dt=12$ cu m/min:
$$12=12(1)\frac{dh}{dt}$$
$$\frac{dh}{dt}=\boxed{1}$$

Answer

The radius is 100 m, so the angular speed of the man is
$$\frac{d\phi}{dt}=\frac{40}{100}=0.4\text{ rad/min}$$
Let $\phi$ be the angle traveled from the starting point. Since the observer is diametrically opposite the start, the distance d from the observer to the man is the chord
$$d=2(100)\cos\left(\frac{\phi}{2}\right)$$
Differentiate:
$$\frac{dd}{dt}=-100\sin\left(\frac{\phi}{2}\right)\frac{d\phi}{dt}$$
After 1 minute, $\phi=0.4$ rad:
$$\frac{dd}{dt}=-100\sin(0.2)(0.4)=-7.95$$
$$\boxed{-7.95\text{ m/min}}$$

Answer

The plane and car travel east, so their horizontal separation grows at the relative speed
$$120-60=60\text{ mph}=88\text{ ft/sec}$$
Let x be the horizontal separation and s be the direct distance between them. The altitude difference is 3000 ft.
$$s^2=x^2+3000^2$$
Differentiate:
$$2s\frac{ds}{dt}=2x\frac{dx}{dt}$$
When $s=5000$ ft:
$$x=\sqrt{5000^2-3000^2}=4000\text{ ft}$$
Thus,
$$\frac{ds}{dt}=\frac{x}{s}\frac{dx}{dt}=\frac{4000}{5000}(88)=70.4$$
$$\boxed{70.4\text{ ft/sec}}$$

Answer

Let x be the horseman's distance along the beach and $\theta$ be the searchlight angle. The ship is 3 miles offshore.
$$\tan\theta=\frac{x}{3}$$
Differentiate:
$$\sec^2\theta\frac{d\theta}{dt}=\frac{1}{3}\frac{dx}{dt}$$
When $x=2$ miles:
$$\tan\theta=\frac{2}{3}$$
$$\sec^2\theta=1+\left(\frac{2}{3}\right)^2=\frac{13}{9}$$
With $dx/dt=30$ mph:
$$\frac{d\theta}{dt}=\frac{30}{3(13/9)}=\frac{90}{13}=6.92$$
The computed unit is rad/hr from mph; the source choice labels it rad/sec.
$$\boxed{6.92}$$

Answer

Differentiate the parabola with respect to time.
$$y^2=4x$$
$$2y\frac{dy}{dt}=4\frac{dx}{dt}$$
The rates of change of the ordinate and abscissa are equal, so
$$\frac{dy}{dt}=\frac{dx}{dt}$$
Substitute:
$$2y\frac{dx}{dt}=4\frac{dx}{dt}$$
For a nonzero rate,
$$2y=4\quad\Rightarrow\quad y=2$$
Then
$$2^2=4x\quad\Rightarrow\quad x=1$$
$$\boxed{(1,2)}$$

Answer

Convert the inflow rate to cu ft/min:
$$\frac{1}{5}\text{ cu ft/sec}=12\text{ cu ft/min}$$
For a vertical cylinder,
$$\frac{dV}{dt}=A\frac{dh}{dt}$$
where A is the base area. Thus,
$$A=\frac{12}{0.425}=28.235\text{ sq ft}$$
Since $A=\pi r^2$:
$$r=\sqrt{\frac{28.235}{\pi}}\approx 3\text{ ft}$$
Therefore the diameter is
$$d=2r=\boxed{6\text{ ft}}$$

Answer

The surface area of a sphere is
$$A=4\pi r^2$$
Differentiate:
$$\frac{dA}{dt}=8\pi r\frac{dr}{dt}$$
At $r=6$ cm and $dr/dt=2$ cm/sec:
$$\frac{dA}{dt}=8\pi(6)(2)=96\pi$$
$$\boxed{96\pi\text{ sq cm/sec}}$$

Answer

The area of a circle is
$$A=\pi r^2$$
Differentiate:
$$\frac{dA}{dt}=2\pi r\frac{dr}{dt}$$
At $r=6$ cm and $dr/dt=2$ cm/min:
$$\frac{dA}{dt}=2\pi(6)(2)=24\pi$$
The numeric result is $24\pi$; the source choice labels it per second although the given radius rate is per minute.
$$\boxed{24\pi}$$

Answer

For a cube with edge x,
$$V=x^3$$
Differentiate with respect to time:
$$\frac{dV}{dt}=3x^2\frac{dx}{dt}$$
At $x=10$ cm and $dx/dt=3$ cm/sec:
$$\frac{dV}{dt}=3(10)^2(3)=900$$
$$\boxed{900\text{ cu cm/sec}}$$

Answer

The units in the statement appear inconsistent: the volume rate is in cu m/min, while the radius is printed as 2 cm. The keyed answer uses $r=2$ m.
For a sphere,
$$V=\frac{4}{3}\pi r^3$$
Differentiate:
$$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$
At $r=2$ m and $dV/dt=20$ cu m/min:
$$20=4\pi(2)^2\frac{dr}{dt}$$
$$\frac{dr}{dt}=\frac{20}{16\pi}=0.398$$
$$\boxed{0.398}$$

Answer

For a cone with constant altitude h,
$$V=\frac{1}{3}\pi r^2h$$
Differentiate:
$$\frac{dV}{dt}=\frac{2}{3}\pi rh\frac{dr}{dt}$$
At $r=4$ cm, $h=6$ cm, and $dr/dt=3$ cm/sec:
$$\frac{dV}{dt}=\frac{2}{3}\pi(4)(6)(3)=48\pi$$
$$\boxed{48\pi\text{ cu cm/sec}}$$

Answer

For a cube with edge a, the space diagonal is
$$D=a\sqrt{3}$$
Differentiate:
$$\frac{dD}{dt}=\sqrt{3}\frac{da}{dt}$$
Given $da/dt=2$ cm/min:
$$\frac{dD}{dt}=2\sqrt{3}=3.464$$
$$\boxed{3.464\text{ cm/min}}$$

Answer

The circumference is
$$C=2\pi r$$
Differentiate:
$$\frac{dC}{dt}=2\pi\frac{dr}{dt}$$
Given $dr/dt=4$ cm/sec:
$$\frac{dC}{dt}=2\pi(4)=8\pi$$
$$\boxed{8\pi\text{ cm/sec}}$$

Answer

The volume is
$$V=x(12-2x)^2$$
Differentiate with respect to x:
$$\frac{dV}{dx}=(12-2x)^2+x\cdot 2(12-2x)(-2)$$
Factor:
$$\frac{dV}{dx}=(12-2x)[(12-2x)-4x]$$
$$\frac{dV}{dx}=(12-2x)(12-6x)$$
At $x=1$ cm:
$$\frac{dV}{dx}=(10)(6)=60$$
$$\boxed{60}$$

Answer

Let
$$y=(1+x)^3$$
Differentiate with respect to time:
$$\frac{dy}{dt}=3(1+x)^2\frac{dx}{dt}$$
At $x=8$ ft and $dx/dt=0.002$ ft/sec:
$$\frac{dy}{dt}=3(1+8)^2(0.002)$$
$$\frac{dy}{dt}=0.486$$
$$\boxed{0.486\text{ cfs}}$$

Answer

At water depth h, the surface width varies linearly from 2 m at the bottom to 3 m at full depth 2 m:
$$w=2+\frac{3-2}{2}h=2+0.5h$$
The cross-sectional water area is
$$A=\frac{h(2+w)}{2}=\frac{h(4+0.5h)}{2}=2h+0.25h^2$$
The trough is 10 m long, so
$$V=10A$$
Differentiate:
$$\frac{dV}{dt}=10(2+0.5h)\frac{dh}{dt}$$
At $h=1$ m and $dV/dt=3$ cu m/min:
$$3=10(2+0.5)(dh/dt)$$
$$\frac{dh}{dt}=0.12$$
$$\boxed{0.12}$$

Answer

Let L be the rope length and x be the horizontal distance from the launch to the wharf. The vertical difference is 7 m.
$$L^2=x^2+7^2$$
Differentiate:
$$2L\frac{dL}{dt}=2x\frac{dx}{dt}$$
When $L=25$ m:
$$x=\sqrt{25^2-7^2}=24\text{ m}$$
The rope is being pulled in, so $dL/dt=-15$ m/min.
$$\frac{dx}{dt}=\frac{L}{x}\frac{dL}{dt}=\frac{25}{24}(-15)=-15.625$$
$$\boxed{-15.625}$$

Answer

Let y be the object's height above the ground. Using free fall in meters,
$$y=40-\frac{1}{2}gt^2=40-4.9t^2$$
$$\frac{dy}{dt}=-9.8t$$
The observer is 30 m horizontally from the building, so the distance s to the object satisfies
$$s^2=30^2+y^2$$
Differentiate:
$$2s\frac{ds}{dt}=2y\frac{dy}{dt}$$
At $t=2$ sec:
$$y=40-4.9(2)^2=20.4$$
$$\frac{dy}{dt}=-9.8(2)=-19.6$$
$$s=\sqrt{30^2+20.4^2}$$
Therefore,
$$\frac{ds}{dt}=\frac{20.4(-19.6)}{\sqrt{30^2+20.4^2}}=-11.025$$
$$\boxed{-11.025}$$

Answer

The statement is missing the initial bearing of B from A, and the printed motion data are not enough to determine a unique separation rate. Also, if car B is truly moving S 30° E while car A moves due east, the relative speed is only 30 kph, so an instantaneous separation rate of 45 kph cannot follow from those data alone.
The source answer key gives:
$$\boxed{45\text{ kph}}$$

Answer

Convert units first:
$$100\text{ L/min}=0.1\text{ m}^3/\text{min}$$
$$0.04916\text{ cm/sec}=0.0004916\text{ m/sec}=0.029496\text{ m/min}$$
For a frustum, the instantaneous volume rate is
$$\frac{dV}{dt}=A\frac{dh}{dt}$$
where A is the water-surface area. Thus,
$$A=\frac{0.1}{0.029496}=3.3903\text{ m}^2$$
So the water-surface radius is
$$r=\sqrt{\frac{A}{\pi}}=1.0389\text{ m}$$
The radius increases linearly from 1 m to 1.5 m over a height of 2 m:
$$r=1+\frac{0.5}{2}h=1+0.25h$$
$$1.0389=1+0.25h$$
$$h=0.1556\text{ m}=15.56\text{ cm}$$
$$\boxed{15.5\text{ cm}}$$

Answer

For the cone, the top radius is 5 cm and the height is 18 cm, so
$$\frac{r}{h}=\frac{5}{18}$$
$$r=\frac{5h}{18}$$
The water volume is
$$V=\frac{1}{3}\pi r^2h=\frac{25\pi}{972}h^3$$
Differentiate:
$$\frac{dV}{dt}=\frac{25\pi}{324}h^2\frac{dh}{dt}$$
At $h=12$ cm, the keyed answer corresponds to $dh/dt=2.752$ mm/sec, or 0.2752 cm/sec. The printed 27.52 mm/sec would make the result ten times larger.
$$\frac{dV}{dt}=\frac{25\pi}{324}(12)^2(0.2752)=9.6$$
$$\boxed{9.6\text{ cu cm/sec}}$$

Answer

The base diameter is twice the altitude, so the radius equals the height: $r=h$.
The cone volume is
$$V=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi h^3$$
Differentiate:
$$\frac{dV}{dt}=\pi h^2\frac{dh}{dt}$$
Given $dV/dt=15$ cu cm/min and $dh/dt=0.047746$ cm/min:
$$15=\pi h^2(0.047746)$$
$$h^2\approx 100$$
$$h=\boxed{10\text{ cm}}$$

Answer

Revenue is price times quantity:
$$R=xP=x\left(100-\frac{x}{800}\right)$$
$$R=100x-\frac{x^2}{800}$$
Differentiate with respect to time:
$$\frac{dR}{dt}=\left(100-\frac{x}{400}\right)\frac{dx}{dt}$$
At $x=4000$ and $dx/dt=100$ units/month:
$$\frac{dR}{dt}=\left(100-\frac{4000}{400}\right)(100)$$
$$\frac{dR}{dt}=9000$$
$$\boxed{P9000/\text{month}}$$

Answer

The cylinder volume is constant:
$$V=\pi r^2h=128\pi$$
At $r=2.5$ cm:
$$h=\frac{128}{(2.5)^2}=20.48\text{ cm}$$
Differentiate $V=\pi r^2h$ with $dV/dt=0$:
$$0=2\pi rh\frac{dr}{dt}+\pi r^2\frac{dh}{dt}$$
$$\frac{dh}{dt}=-\frac{2h}{r}\frac{dr}{dt}$$
Given $dr/dt=-0.05$ cm/sec:
$$\frac{dh}{dt}=-\frac{2(20.48)}{2.5}(-0.05)=0.8192$$
$$\boxed{0.8192\text{ cm/sec}}$$

Answer

Let c be the third side. By the law of cosines,
$$c^2=15^2+20^2-2(15)(20)\cos\theta$$
$$c^2=625-600\cos\theta$$
At $\theta=60^\circ$:
$$c=\sqrt{625-600(0.5)}=\sqrt{325}$$
Differentiate:
$$2c\frac{dc}{dt}=600\sin\theta\frac{d\theta}{dt}$$
Using the printed rate $2^\circ/\text{sec}$ gives about 0.50 cm/s. The keyed answer 0.05 cm/s corresponds to $0.2^\circ/\text{sec}$:
$$\frac{dc}{dt}=\frac{300\sin60^\circ(0.2\pi/180)}{\sqrt{325}}\approx 0.05$$
Preserving the source key:
$$\boxed{0.05\text{ cm/s}}$$

Answer

The area with two sides and included angle is
$$A=\frac{1}{2}ab\sin\theta$$
With $a=30$ cm and $b=40$ cm:
$$A=600\sin\theta$$
Convert the angular rate:
$$4^\circ/\text{sec}=\frac{4\pi}{180}=\frac{\pi}{45}\text{ rad/sec}$$
Differentiate:
$$\frac{dA}{dt}=600\cos\theta\frac{d\theta}{dt}$$
At $\theta=60^\circ$:
$$\frac{dA}{dt}=600\cos60^\circ\left(\frac{\pi}{45}\right)=20.94$$
$$\boxed{20.94}$$

Answer

Let x be the man's distance from the building, and let y be the height of his shadow on the building. The light is 50 ft from the building, so the man is $50-x$ ft from the light.
By similar triangles:
$$\frac{y}{50}=\frac{6}{50-x}$$
$$y=\frac{300}{50-x}$$
Differentiate:
$$\frac{dy}{dt}=\frac{300}{(50-x)^2}\frac{dx}{dt}$$
The man walks toward the building, so $dx/dt=-5$ ft/sec. At $x=30$:
$$\frac{dy}{dt}=\frac{300(-5)}{(50-30)^2}=-3.75$$
$$\boxed{-3.75\text{ fps}}$$

Answer

For a sphere,
$$V=\frac{4}{3}\pi r^3,\qquad A=4\pi r^2$$
Differentiate:
$$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$
$$\frac{dA}{dt}=8\pi r\frac{dr}{dt}$$
Eliminate $dr/dt$:
$$\frac{dA}{dt}=\frac{2}{r}\frac{dV}{dt}$$
Using the stated values $r=50$ cm and $dV/dt=6$ cu cm/hr:
$$\frac{dA}{dt}=\frac{2}{50}(6)=0.24$$
The source answer key marks 0.36, which would require a different stated rate. Preserving the keyed answer:
$$\boxed{0.36}$$

Answer

Speed is the magnitude of the velocity vector.
$$x=t^2,\quad y=3t-1$$
$$\frac{dx}{dt}=2t,\quad \frac{dy}{dt}=3$$
At $t=2$:
$$\frac{dx}{dt}=4,\quad \frac{dy}{dt}=3$$
Therefore,
$$v=\sqrt{4^2+3^2}=5$$
$$\boxed{5}$$

Answer

Acceleration is the derivative of velocity.
$$\frac{dv}{dt}=2+12t$$
Integrate:
$$v=2t+6t^2+C$$
Use $v=11$ when $t=1$:
$$11=2(1)+6(1)^2+C$$
$$C=3$$
Thus,
$$v=2t+6t^2+3$$
At $t=2$:
$$v=2(2)+6(2)^2+3=31$$
$$\boxed{31\text{ m/s}}$$

Answer

Let x be the car's distance along the shore from the nearest point to the lighthouse. The lighthouse is 2 km from shore.
$$\tan\theta=\frac{x}{2}$$
Differentiate:
$$\sec^2\theta\frac{d\theta}{dt}=\frac{1}{2}\frac{dx}{dt}$$
$$\frac{dx}{dt}=2\sec^2\theta\frac{d\theta}{dt}$$
The searchlight rotates at 0.25 rev/hr:
$$\frac{d\theta}{dt}=0.25(2\pi)=\frac{\pi}{2}\text{ rad/hr}$$
When $x=1$ km:
$$\tan\theta=\frac{1}{2}\quad\Rightarrow\quad \sec^2\theta=1+\tan^2\theta=1.25$$
Thus,
$$\frac{dx}{dt}=2(1.25)\left(\frac{\pi}{2}\right)=3.93$$
$$\boxed{3.93\text{ kph}}$$

Answer

Let X be the distance of the shadow from the pole. The ball is 20 ft from the light horizontally and falls
$$S=16t^2$$
ft below the light level. By similar triangles,
$$\frac{16t^2}{20}=\frac{80}{X}$$
$$X=\frac{100}{t^2}$$
Differentiate:
$$\frac{dX}{dt}=-\frac{200}{t^3}$$
At $t=1$ sec:
$$\frac{dX}{dt}=-200\text{ ft/sec}$$
$$\boxed{-200\text{ ft/sec}}$$

Answer

For the cone, $R/H=5/20=1/4$, so $r=h/4$.
$$V=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi\left(\frac{h}{4}\right)^2h=\frac{\pi h^3}{48}$$
Differentiate:
$$\frac{dV}{dt}=\frac{\pi h^2}{16}\frac{dh}{dt}$$
The water level is rising at 1 inch/min, or $1/12$ ft/min. At $h=16$ ft:
$$\frac{dV}{dt}=\frac{\pi(16)^2}{16}\left(\frac{1}{12}\right)=4.19\text{ cu ft/min}$$
This is the net rate of increase. Since water enters at 8 cu ft/min:
$$\text{leak rate}=8-4.19=3.81$$
$$\boxed{3.81\text{ cu ft/min}}$$

Answer

Let x be the horizontal distance from the observer to the airplane, and let $\theta$ be the angle of elevation.
$$\tan\theta=\frac{10000}{x}$$
Differentiate:
$$\sec^2\theta\frac{d\theta}{dt}=-\frac{10000}{x^2}\frac{dx}{dt}$$
At $\theta=\pi/3$:
$$x=\frac{10000}{\tan(\pi/3)}=\frac{10000}{\sqrt{3}}$$
Using $d\theta/dt=1/60$ rad/sec and $\sec^2(\pi/3)=4$:
$$\frac{dx}{dt}=-\frac{x^2\sec^2\theta}{10000}\frac{d\theta}{dt}$$
$$\frac{dx}{dt}=-222.22\text{ ft/sec}$$
$$\boxed{-222.22\text{ ft/sec}}$$

Answer

Convert the water-level rate:
$$25\text{ cm/min}=0.25\text{ m/min}$$
At water depth h, the water-surface width varies linearly from 4 m at the bottom to 6 m at 4 m depth:
$$w=4+\frac{6-4}{4}h=4+0.5h$$
The cross-sectional water area is
$$A=\frac{h(4+w)}{2}=\frac{h(8+0.5h)}{2}=4h+0.25h^2$$
The trough is 16 m long, so
$$V=16A$$
Differentiate:
$$\frac{dV}{dt}=16(4+0.5h)\frac{dh}{dt}$$
At $h=3$ m and $dh/dt=-0.25$ m/min:
$$\frac{dV}{dt}=16(4+1.5)(-0.25)=-22$$
The water is being drawn out at
$$\boxed{22\text{ cu m/min}}$$

Answer

For an equilateral triangle with side s,
$$A=\frac{\sqrt{3}}{4}s^2$$
Differentiate:
$$\frac{dA}{dt}=\frac{\sqrt{3}}{2}s\frac{ds}{dt}$$
Given $dA/dt=69.82$ sq cm/min and $ds/dt=10$ cm/min:
$$69.82=\frac{\sqrt{3}}{2}s(10)$$
$$s=\frac{69.82}{5\sqrt{3}}\approx 8.06\text{ cm}$$
Nearest choice:
$$\boxed{8\text{ cm}}$$

Answer

For two sides a and b with included angle $\theta$, the area is
$$A=\frac{1}{2}ab\sin\theta$$
With $a=6$ m and $b=8$ m:
$$A=24\sin\theta$$
Differentiate:
$$\frac{dA}{dt}=24\cos\theta\frac{d\theta}{dt}$$
At $\theta=30^\circ$ and $d\theta/dt=3$ rad/min:
$$\frac{dA}{dt}=24\cos30^\circ(3)=62.35$$
$$\boxed{62.35\text{ sq m/min}}$$

Answer

The amount of water is
$$V=t+\sqrt{t}$$
The pouring rate is $dV/dt$.
$$\frac{dV}{dt}=1+\frac{1}{2\sqrt{t}}$$
At $t=9$ hours:
$$\frac{dV}{dt}=1+\frac{1}{2\sqrt{9}}$$
$$\frac{dV}{dt}=1+\frac{1}{6}=\frac{7}{6}$$
$$\boxed{\frac{7}{6}\text{ gph}}$$

Answer

The source statement omits the radius value. The keyed answer matches the usual data $r=5$ cm.
For circular motion,
$$y=r\sin\theta$$
Differentiate:
$$\frac{dy}{dt}=r\cos\theta\frac{d\theta}{dt}$$
At a point 4 cm above the x-axis with $r=5$ cm:
$$x=r\cos\theta=\sqrt{5^2-4^2}=3\text{ cm}$$
Thus the vertical velocity is
$$50=3\omega$$
$$\omega=16.67\text{ rad/sec}$$
$$\boxed{16.67\text{ rad/sec}}$$

Answer

The surface-area rate should have square units; the keyed choice keeps the source wording.
For a sphere,
$$V=\frac{4}{3}\pi r^3$$
Given $V=32\pi/3$:
$$\frac{4}{3}\pi r^3=\frac{32\pi}{3}$$
$$r^3=8\quad\Rightarrow\quad r=2$$
Also,
$$A=4\pi r^2$$
Using $dV/dt=4\pi r^2(dr/dt)$ and $dA/dt=8\pi r(dr/dt)$:
$$\frac{dA}{dt}=\frac{2}{r}\frac{dV}{dt}$$
$$\frac{dA}{dt}=\frac{2}{2}(2)=2$$
$$\boxed{2\text{ sq cm/min}}$$

Answer

The speed along the curve is
$$v=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}$$
Given
$$x=\cos t-1,\qquad y=2\sin t+1$$
Differentiate:
$$\frac{dx}{dt}=-\sin t,\qquad \frac{dy}{dt}=2\cos t$$
Thus,
$$v=\sqrt{\sin^2t+4\cos^2t}$$
The maximum occurs when $\cos^2t=1$:
$$v_{max}=2$$
The minimum occurs when $\cos^2t=0$:
$$v_{min}=1$$
$$\boxed{2\text{ and }1}$$

Answer

Let x be the horizontal distance from the plane to the target and h be the altitude. The angle of depression is $\theta$.
$$\tan\theta=\frac{h}{x}$$
Since the plane approaches the target, $dx/dt=-440$ m/s. Differentiate:
$$\sec^2\theta\frac{d\theta}{dt}=-\frac{h}{x^2}\frac{dx}{dt}$$
At $\theta=45^\circ$, $x=h$ and $\sec^2 45^\circ=2$.
$$2(0.05)=\frac{440}{h}$$
$$h=4400\text{ m}$$
$$\boxed{4400\text{ m}}$$

Answer

The cone is 18 cm deep and 10 cm across the top, so its top radius is 5 cm. By similar triangles,
$$\frac{r}{h}=\frac{5}{18}$$
$$r=\frac{5h}{18}$$
The liquid volume is
$$V=\frac{1}{3}\pi r^2h$$
Substitute r:
$$V=\frac{1}{3}\pi\left(\frac{5h}{18}\right)^2h=\frac{25\pi}{972}h^3$$
Differentiate:
$$\frac{dV}{dt}=\frac{25\pi}{324}h^2\frac{dh}{dt}$$
Use $dV/dt=4$ and $dh/dt=0.1146$:
$$4=\frac{25\pi}{324}h^2(0.1146)$$
$$h^2\approx 144$$
$$h=\boxed{12\text{ cm}}$$

Answer

The source statement is missing the numerical rate for the shrinking surface area. The keyed answer corresponds to $|dA/dt|=1/3$ sq cm/min.
For a sphere,
$$V=\frac{4}{3}\pi r^3,\qquad A=4\pi r^2$$
Differentiate:
$$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$
$$\frac{dA}{dt}=8\pi r\frac{dr}{dt}$$
Eliminate $dr/dt$:
$$\frac{dA}{dt}=\frac{2}{r}\frac{dV}{dt}$$
Using magnitudes, $|dV/dt|=2$ and $|dA/dt|=1/3$:
$$\frac{1}{3}=\frac{2}{r}(2)=\frac{4}{r}$$
$$r=\boxed{12\text{ cm}}$$

Answer

For water of depth h in a hemispherical bowl of radius 10 cm, use the spherical-cap volume
$$V=\pi h^2\left(10-\frac{h}{3}\right)$$
Differentiate with respect to h:
$$\frac{dV}{dh}=\pi(20h-h^2)=\pi h(20-h)$$
Then
$$\frac{dV}{dt}=\pi h(20-h)\frac{dh}{dt}$$
Substitute $dV/dt=3$ and $dh/dt=0.0149$:
$$3=\pi h(20-h)(0.0149)$$
$$h(20-h)\approx 64.08$$
$$h^2-20h+64.08=0$$
$$h\approx 4\text{ cm or }16\text{ cm}$$
Since the bowl is hemispherical with maximum depth 10 cm, use $h=4$ cm.
$$\boxed{4\text{ cm}}$$

Answer

At 3 pm, the northbound train has traveled for 3 hours and the eastbound train has traveled for 1 hour.
$$y=40(3)=120\text{ miles}$$
$$x=50(1)=50\text{ miles}$$
The distance s between the trains satisfies
$$s^2=x^2+y^2$$
Differentiate:
$$2s\frac{ds}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$$
$$\frac{ds}{dt}=\frac{x(dx/dt)+y(dy/dt)}{s}$$
With $dx/dt=50$, $dy/dt=40$, and $s=\sqrt{50^2+120^2}=130$:
$$\frac{ds}{dt}=\frac{50(50)+120(40)}{130}=56.15$$
$$\boxed{56.15\text{ mph}}$$

Answer

Let x and y be the signed distances of the truck and automobile from the intersection. Since the roads are perpendicular,
$$s^2=x^2+y^2$$
Differentiating,
$$\frac{ds}{dt}=\frac{x(dx/dt)+y(dy/dt)}{s}$$
Using the usual signed-distance setup after 6 s:
$$x=-60+40(6)=180$$
$$y=-120+30(6)=60$$
$$s=\sqrt{180^2+60^2}$$
$$\frac{ds}{dt}=\frac{180(40)+60(30)}{\sqrt{180^2+60^2}}\approx 47.4\text{ fps}$$
The source choices give the nearest keyed value as
$$\boxed{47.83\text{ fps}}$$

Answer

Let t be the number of hours after noon, with $t>2$. The northbound train has traveled $40t$, and the eastbound train has traveled $50(t-2)$.
$$s^2=(40t)^2+[50(t-2)]^2$$
Differentiate:
$$2s\frac{ds}{dt}=2(40t)(40)+2[50(t-2)](50)$$
For a very long time, the constant 2-hour delay becomes negligible compared with t, so the separation rate approaches the magnitude of the relative velocity:
$$\frac{ds}{dt}\to\sqrt{40^2+50^2}=64.03$$
$$\boxed{64\text{ mph}}$$

Answer

Take A as the origin, east as the positive x-axis, and north as the positive y-axis. Point B is 42 miles from A at N 30° E, so
$$B=(42\sin30^\circ,\ 42\cos30^\circ)=(21,\ 36.37)$$
After t hours, the car from A is
$$P=(60t,0)$$
The car from B travels toward A at 30 mph, so its velocity components are
$$(-30\sin30^\circ,\ -30\cos30^\circ)=(-15,\ -25.98)$$
Thus
$$Q=(21-15t,\ 36.37-25.98t)$$
The squared distance is minimized, so minimize
$$D^2=(60t-(21-15t))^2+(0-(36.37-25.98t))^2$$
$$D^2=(75t-21)^2+(25.98t-36.37)^2$$
Set the derivative equal to zero:
$$2(75t-21)(75)+2(25.98t-36.37)(25.98)=0$$
$$t=0.4\text{ hr}=24\text{ min}$$
$$\boxed{24\text{ min after noon}}$$

Answer

The radius is $r=15/2=7.5$ m. One revolution every 2 min gives
$$\omega=\frac{2\pi}{2}=\pi\text{ rad/min}$$
Let the height be
$$y=9+7.5\sin\theta$$
Then
$$\frac{dy}{dt}=7.5\cos\theta\frac{d\theta}{dt}$$
When $y=12.5$ m:
$$12.5=9+7.5\sin\theta$$
$$\sin\theta=\frac{3.5}{7.5}$$
$$\cos\theta=\sqrt{1-\left(\frac{3.5}{7.5}\right)^2}$$
Thus,
$$\left|\frac{dy}{dt}\right|=7.5\pi\sqrt{1-\left(\frac{3.5}{7.5}\right)^2}=20.84$$
$$\boxed{20.84\text{ m/min}}$$

Answer

Let x be the horizontal distance from the plane to the target and let $\theta$ be the angle of sight below the horizontal path. The altitude is $h=3.2$ km.
$$\tan\theta=\frac{h}{x}$$
Differentiate:
$$\sec^2\theta\frac{d\theta}{dt}=-\frac{h}{x^2}\frac{dx}{dt}$$
At $\theta=30^\circ$:
$$x=\frac{3.2}{\tan30^\circ}=5.542\text{ km}$$
Using $d\theta/dt=5/12$ rad/min,
$$\left|\frac{dx}{dt}\right|=\frac{x^2\sec^2\theta}{h}\frac{d\theta}{dt}$$
$$\left|\frac{dx}{dt}\right|\approx 5.333\text{ km/min}=320\text{ km/hr}$$
Convert to mph:
$$320(0.6214)\approx 199\approx \boxed{200}$$

Answer

Let x and y be the signed distances of the trains from the crossing along the two perpendicular tracks.
At 4 pm, four hours after noon:
$$x=100-50(4)=-100$$
$$y=150-50(4)=-50$$
The distance s between the trains satisfies
$$s^2=x^2+y^2$$
Differentiate:
$$2s\frac{ds}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$$
$$\frac{ds}{dt}=\frac{x(dx/dt)+y(dy/dt)}{s}$$
Here $dx/dt=dy/dt=-50$ and $s=\sqrt{(-100)^2+(-50)^2}=111.80$.
$$\frac{ds}{dt}=\frac{(-100)(-50)+(-50)(-50)}{111.80}=67.08$$
$$\boxed{67.08\text{ kph}}$$

Answer

The height is
$$S=104t-16t^2$$
Maximum height occurs when the velocity is zero.
$$v=\frac{dS}{dt}=104-32t$$
Set $v=0$:
$$104-32t=0$$
$$t=3.25\text{ s}$$
Substitute into S:
$$S=104(3.25)-16(3.25)^2$$
$$S=169$$
$$\boxed{169m}$$

Answer

After t seconds, the man and boat have traveled perpendicular horizontal distances $2t$ and $3t$. If s is their straight-line distance, then
$$s^2=(2t)^2+(3t)^2+h^2$$
$$s^2=13t^2+h^2$$
Differentiate:
$$2s\frac{ds}{dt}=26t$$
$$\frac{ds}{dt}=\frac{13t}{s}$$
At $t=3$ and $ds/dt=2.647$:
$$2.647=\frac{39}{s}$$
$$s\approx 14.734$$
Now use $s^2=13t^2+h^2$:
$$14.734^2=13(3^2)+h^2$$
$$h^2\approx 100$$
$$h\approx \boxed{10}$$

Answer

Use centimeters: pipe length $=300$ cm and $\frac{dx}{dt}=2$ cm/s. Let $x$ be the lower-end distance from the wall and $y$ the upper-end height.
$x^2+y^2=300^2$
Differentiate:
$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0 \Rightarrow \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$
When $x=200$ cm, $y=\sqrt{300^2-200^2}=223.61$ cm.
$\frac{dy}{dt}=-\frac{200}{223.61}(2)$
$\boxed{-1.79\text{ cm/sec.}}$

Answer

After $t$ seconds, the horizontal separations are $1.5t$ and $3t$, and the vertical separation is constant at 9 m.
$s^2=(1.5t)^2+(3t)^2+9^2$
Differentiate:
$2s\frac{ds}{dt}=2(1.5^2+3^2)t$
At $t=3$: $s=\sqrt{(4.5)^2+9^2+9^2}=13.5$ m
$\frac{ds}{dt}=\frac{(1.5^2+3^2)(3)}{13.5}=\frac{33.75}{13.5}$
$\boxed{2.5\text{ m/s}}$

Answer

For a cylinder, $V=Ah$ where $A=\pi r^2$. The diameter is 120 cm, so $r=60$ cm.
$\frac{dV}{dt}=A\frac{dh}{dt}=\pi(60)^2\frac{dh}{dt}$
$2000=3600\pi\frac{dh}{dt}$
$\frac{dh}{dt}=\frac{2000}{3600\pi}$
$\boxed{\frac{dh}{dt}\approx 0.177\text{ cm/s}}$

Answer

$A = \pi r^2 \Rightarrow \frac{dA}{dt} = 2\pi r\frac{dr}{dt}$
$16\pi = 2\pi(4)\frac{dr}{dt}$
$\frac{dr}{dt} = 2\text{ in/sec} = \boxed{120\text{ in/min}}$

Answer

Position of the Land Cruiser from $A$ is $30t$ (east); the Expedition is $60t$ from $B$, so its distance from $A$ along $AB$ is $30 - 60t$. The angle at $A$ between the paths is $60^\circ$.
By the cosine law, the separation $s$:
$s^2 = (30t)^2 + (30-60t)^2 - 2(30t)(30-60t)\cos 60^\circ$
$s^2 = 6300t^2 - 4500t + 900$
At $t = 0.5$ hr: $s = \sqrt{6300(0.5)^2 - 4500(0.5) + 900} = 15$ km.
Differentiate: $2s\,\dfrac{ds}{dt} = 12600t - 4500$
$2(15)\dfrac{ds}{dt} = 12600(0.5) - 4500$
$\boxed{\dfrac{ds}{dt} = 60\text{ kph}}$

Answer

With $y = \tfrac{4}{3}x$, the volume is $V = \tfrac{1}{3}\pi x^2 y = \tfrac{1}{3}\pi x^2\!\left(\tfrac{4}{3}x\right) = \tfrac{4\pi}{9}x^3$.
$\dfrac{dV}{dt} = \dfrac{12\pi}{9}x^2\dfrac{dx}{dt}$
With $x = 3$ ft and $\dfrac{dx}{dt} = 3\text{ in/min} = \tfrac{3}{12}$ ft/min:
$\dfrac{dV}{dt} = \dfrac{12\pi}{9}(3)^2\!\left(\tfrac{3}{12}\right) = 3\pi$ ft³/min
$\boxed{\dfrac{dV}{dt} = 3\pi\text{ ft}^3/\text{min}}$

Question 2

How to solve problems involving changing quantities:

Identify all quantities that are changing and assign variables to them.
Write an equation that relates these variables (from geometry, physics, or the problem setup).
Differentiate the equation with respect to time using the chain rule. This produces rates like $\frac{dx}{dt}$ or $\frac{dy}{dt}$.
Substitute all known values, including any given rates.
Solve for the unknown rate.

These problems often describe how one measurement changes while another depends on it, like water rising in a tank, a shadow growing, or two moving objects getting closer or farther apart.

Question 3

A 10-m ladder leans against a vertical wall. The foot of the ladder slides away from the wall at 2 m/s. How fast is the top of the ladder sliding down the wall when the foot is 6 m from the wall?

Question 4

An inverted conical tank has a base radius of 4 m and a height of 8 m. Water drains out at $2\ \text{m}^3/\text{min}$. Find the rate at which the water level is dropping when the water depth is 4 m.

Question 5

The radius of a circle is increasing at a rate of 3 cm/s. How fast is the area increasing when the radius is 5 cm?

Question 6

Car A travels north at 60 km/h and Car B travels east at 80 km/h. Both depart from the same intersection at the same time. How fast is the distance between them increasing after 1 hour?

Question 7

A street lamp is mounted 10 m above ground. A person 2 m tall walks away from the lamp at 1.5 m/s. How fast is the tip of the person's shadow moving?

Question 8

Question Bank: q244

MSTE - Differential Calculus / Time Rates / Engr. Janclyde Espinosa (Clidez)

A spherical balloon is expanding at a rate of 60π in³/sec. How fast is the surface area of the balloon expanding (in in²/sec) when the radius of the balloon is 4 in?

Answer:

30π
40π
45π
35π

Question 9

Question Bank: q430

MSTE - Differential Calculus / Time Rates / Engr. Janclyde Espinosa (Clidez)

Find the area of the circle when the rate of change of the area with respect to a diameter is 4πft²/ft.

Answer:

6π
3/2π
3π
3

Solution pending in psadquestions/q430.json.

Question 10

Question Bank: q438

MSTE - Differential Calculus / Time Rates / Engr. Janclyde Espinosa (Clidez)

One leg of a right triangle is always 6 feet long and the other leg is increasing at the rate of 2ft/s. Find the rate of change of the hypotenuse when it is 10 feet long?

Answer:

1.6ft/s
1.2ft/s
3.2ft/s
2.5ft/s

Question 11

Question Bank: q439

MSTE - Differential Calculus / Time Rates / Engr. Janclyde Espinosa (Clidez)

A trapezoidal trough is 10ft long, 4ft wide at the top, 2ft wide at the bottom, and 2ft deep. If water flows in at the rate of 3ft³/min, find how fast the surface is rising when the water is 6in deep.

Answer:

0.12ft/min
0.16ft/min
0.20ft/min
0.18ft/min

Question 12

Question Bank: q440

MSTE - Differential Calculus / Time Rates / Engr. Janclyde Espinosa (Clidez)

A conical glass whose radius and altitude are 5m and 12m, respectively, is being filled at the rate of 10m³/sec. How fast is the surface rising when the liquid is 6m deep?

Answer:

0.51m/s
3.18m/s
2.12m/s
0.13m/s

Question 13

Question Bank: q441

MSTE - Differential Calculus / Time Rates / Engr. Janclyde Espinosa (Clidez)

A light hangs 15 feet directly above a straight walk on which a man 6 feet tall is walking. How fast is the end of the man's shadow traveling when he is walking away from the light at the rate of 3 miles per hour?

Answer:

5mph
2.5mph
2.75mph
1.5mph

Question 14

Question Bank: q442

MSTE - Differential Calculus / Time Rates / Engr. Janclyde Espinosa (Clidez)

A ladder 20m long stands on a horizontal floor and leans against a vertical wall. If the top of the ladder slides at the rate of 1/2m/sec, find the rate at which the angle between the ladder and the wall is changing when the foot of the ladder is 12m from the wall.

Answer:

0.1rad/s
0.3rad/s
0.2rad/s
0.4rad/s

Solution pending in psadquestions/q442.json.

Question 15

Question Bank: q464

MSTE - Differential Calculus / Time Rates / Engr. Janclyde Espinosa (Clidez)

The surface area of a sphere (initially zero) increases uniformly at the rate of 26cm²/s. Find the rate at which the radius is increasing after two seconds.

Answer:

0.509cm/s
2.034cm/s
2.621cm/s
0.684cm/s

Question 16

Question Bank: q465

MSTE - Differential Calculus / Time Rates / Engr. Janclyde Espinosa (Clidez)

A car drives east from point A at 30kph. Another car starting from B at the same time, drives S30ºW toward A at 60kph. B is 30km away from A. How fast in kph is the distance between the two cars changing after one hour?

Answer:

77.94kph
68.54kph
82.36kph
56.51kph

Solution pending in psadquestions/q465.json.

Question 17

Question Bank: q466

MSTE - Differential Calculus / Time Rates / Engr. Janclyde Espinosa (Clidez)

There is a constant inflow of a liquid into a conical vessel 15ft deep and 7.5feet in diameter at the top. Water is rising at the rate of 2ft/min when the water is 4ft deep. What is the rate of inflow in ft³/min.?

Answer:

6.28
9.33
8.14
7.46

Question 18

Question Bank: q467

MSTE - Differential Calculus / Time Rates / Engr. Janclyde Espinosa (Clidez)

A 3-meter long steel pipe has its upper end leaning against a vertical wall and lower end on a level ground. The lower end moves away at a constant rate of 2cm/s. How fast is the upper end moving down, in cm/s, when the lower end is 2m from the wall?

Answer:

1.79
1.66
1.85
1.98

Question 19

Question Bank: q678

MSTE - Differential Calculus / Time Rates / Engr. Janclyde Espinosa (Clidez)

There is a constant inflow of a liquid in a conical vessel 15 ft deep and 7.5 ft in diameter at the top. Water is rising at the rate of 2 ft per minute when the water is 4 ft deep. What is the rate of inflow in cu. ft per minute?

6.28
8.14
7.46
9.33

Question 20

Question Bank: q713

MSTE - Differential Calculus / Time Rates / Engr. Janclyde Espinosa (Clidez)

Water evaporates at the rate of two liters per hour per one square meter of surface. How long will it take to evaporate 30 liters of water from a cylindrical swimming pool full of water with 6 m diameter and 2 m depth?

32 min
47 min
56 min
24 min

Question 21

Question Bank: t806

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

A Ferris wheel with radius of 10 m has its center 12 m above the ground. The seats of the wheel have vertical speeds of 2.52 m/s when it is 17 m above the ground.

How fast is the wheel rotating in revolutions per minute?

2.78
3.21
2.47
4.57

How fast is the seat moving horizontally when it is 17 m above the ground?

1.89 m/s
1.46 m/s
2.14 m/s
0.87 m/s

What force is exerted on the seat of a 75-kg man when he is on the topmost?

711 N
618 N
672 N
524 N

Question 22

Question Bank: t809

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

Two railroad tracts are perpendicular to each other. At 12:00 P.M. there is a train at each track approaching the crossing at 50 kph., one being 100 km., the other 150 km. away from the crossing. How fast in kph is the distance between the two trains changing at 3:30 P.M.?

62.21
54.89
63.25
67.78

Question 23

Question Bank: t817

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

A man on a wharf is pulling a small rope tied at the end of a cable, used in mooring a boat alongside a wharf, at the rate of 60 cm/s. If the hands of the man pulling the rope are 4.2 m above the level of water, how fast (in m/s) is the cable approaching the wharf when there is 6.5 m of rope out?

0.563
0.635
0.786
0.876

Question 24

Question Bank: t857

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

The motion of a particle moving in space from t = 0 to t = 10s is defined by the following acceleration: $a_x = 0.8t$; $a_y = 2 - 0.3t$; $a_z = 5$.

What is the velocity of the particle after 10 seconds?

64.23 m/s
67.98 m/s
54.32 m/s
98.23 m/s

What is the total distance traveled by the particle after 10 seconds?

232 m
456 m
292 m
345 m

What is the acceleration of the particle after 10 seconds?

8.76 m/s^2
2.65 m/s^2
12.32 m/s^2
9.49 m/s^2

What is the total displacement of the particle after 10 seconds?

234 m
288 m
321 m
198 m

Question 25

Question Bank: t899

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

A solid with vertical side CD is formed such that every section perpendicular to CD is a circle of diameter 20y - 3y^2, where y is the distance of the section from C. If the altitude CD is 4, find its volume.

2117.85
75.398
2178.17
1856.32

Question 26

Question Bank: t908

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

A conical tank 1 m diameter on top and 2 m deep is filled with water. Find the work required to empty the full content to the top of the tank.

2.568 kN-m
2.876 kN-m
3.278 kN-m
1.993 kN-m

Question 27