Higher Order and Implicit Differentiation
$$y' = \frac{dy}{dx}$$
$$y'' = \frac{d^2 y}{dx^2}$$
$$y^{(n)} = \frac{d^n y}{dx^n}$$
Implicit differentiation:
$$\frac{d}{dx}F(x,y) = F_x + F_y \frac{dy}{dx} = 0$$
$$\frac{dy}{dx} = -\frac{F_x}{F_y}$$
Steps in Solving Implicit Differentiation Problems
How to differentiate equations where $y$ is not isolated:
Differentiate both sides of the equation with respect to $x$.
Apply the chain rule whenever you differentiate a term involving $y$, which produces $\frac{dy}{dx}$.
Collect all terms containing $\frac{dy}{dx}$ on one side of the equation.
Factor out $\frac{dy}{dx}$.
Solve for $\frac{dy}{dx}$.
Implicit differentiation is used when the equation cannot easily be solved for $y$, such as circles, ellipses, or equations where $x$ and $y$ are mixed together.
Book Examples for Implicit and Second Derivatives
Example 1. Find $\dfrac{dy}{dx}$ if $y^2 = 4x^2 + 9$.
Show Solution
Differentiate both sides with respect to $x$. Since $y$ is a function of $x$, use the chain rule on $y^2$.
$$ \frac{d}{dx}(y^2) = \frac{d}{dx}(4x^2 + 9) $$
$$ 2y\frac{dy}{dx} = 8x $$
Divide both sides by $2y$.
$$ \boxed{\frac{dy}{dx} = \frac{4x}{y}} $$
Example 2. Find $y'$ if $x^2 + 4xy + 4y^2 = 0$.
Show Solution
Differentiate term by term. The term $4xy$ needs the product rule.
$$ 2x + 4\left(xy' + y\right) + 8yy' = 0 $$
Expand, then collect all terms with $y'$.
$$ 2x + 4xy' + 4y + 8yy' = 0 $$
$$ 4xy' + 8yy' = -2x - 4y $$
$$ (4x + 8y)y' = -(2x + 4y) $$
Divide by $4x + 8y$ and simplify by $2$.
$$ y' = -\frac{2x + 4y}{4x + 8y} $$
$$ \boxed{y' = -\frac{x + 2y}{2x + 4y}} $$
Example 3. Find $y''$ if $x^2 + y^2 = 4$.
Show Solution
First find $y'$ by differentiating implicitly.
$$ 2x + 2yy' = 0 $$
$$ y' = -\frac{x}{y} $$
Now differentiate $y' = -\dfrac{x}{y}$ using the quotient rule.
$$ y'' = -\frac{(1)(y) - x(y')}{y^2} $$
Substitute $y' = -\dfrac{x}{y}$.
$$ y'' = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} $$
$$ = -\frac{y + \frac{x^2}{y}}{y^2}
= -\frac{x^2 + y^2}{y^3} $$
Since $x^2 + y^2 = 4$, replace the numerator by $4$.
$$ \boxed{y'' = -\frac{4}{y^3}} $$
Exercise 2.5 - Implicit Differentiation
Find $\dfrac{dy}{dx}$ by implicit differentiation. Treat $y$ as a function of $x$, so every derivative of $y$ must include $y'$.
Derivative of $x^3 + y^3 - 6xy = 0$
Find $\dfrac{dy}{dx}$ if $x^3 + y^3 - 6xy = 0$.
Show Solution
Differentiate each term with respect to $x$. Use the product rule for $xy$.
$$ 3x^2 + 3y^2y' - 6(xy' + y) = 0 $$
Expand and collect the $y'$ terms.
$$ 3x^2 + 3y^2y' - 6xy' - 6y = 0 $$
$$ (3y^2 - 6x)y' = 6y - 3x^2 $$
Divide by $3y^2 - 6x$, then simplify by $3$.
$$ \boxed{\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}} $$
Derivative of $x^2 + xy^2 + y^2 = 1$
Find $\dfrac{dy}{dx}$ if $x^2 + xy^2 + y^2 = 1$.
Show Solution
The middle term $xy^2$ is a product: first factor $x$, second factor $y^2$.
$$ 2x + \frac{d}{dx}(xy^2) + 2yy' = 0 $$
$$ \frac{d}{dx}(xy^2) = x(2yy') + y^2(1) = 2xyy' + y^2 $$
Substitute and collect the $y'$ terms.
$$ 2x + 2xyy' + y^2 + 2yy' = 0 $$
$$ (2xy + 2y)y' = -2x - y^2 $$
$$ \boxed{\frac{dy}{dx} = -\frac{2x + y^2}{2y(x+1)}} $$
Derivative of $\sqrt{x+y} + xy = 21$
Find $\dfrac{dy}{dx}$ if $\sqrt{x+y} + xy = 21$.
Show Solution
Differentiate $\sqrt{x+y}$ as $(x+y)^{1/2}$, then use the chain rule.
$$ \frac{1}{2\sqrt{x+y}}(1+y') + (xy' + y) = 0 $$
Multiply the whole equation by $2\sqrt{x+y}$ to remove the fraction.
$$ 1 + y' + 2\sqrt{x+y}(xy' + y) = 0 $$
Expand and collect the $y'$ terms.
$$ 1 + y' + 2x\sqrt{x+y}\,y' + 2y\sqrt{x+y} = 0 $$
$$ (1 + 2x\sqrt{x+y})y' = -(1 + 2y\sqrt{x+y}) $$
$$ \boxed{\frac{dy}{dx} = -\frac{1 + 2y\sqrt{x+y}}{1 + 2x\sqrt{x+y}}} $$
Derivative of $\sqrt{x} + \sqrt{y} = \sqrt{a}$
Find $\dfrac{dy}{dx}$ if $\sqrt{x} + \sqrt{y} = \sqrt{a}$, where $a$ is constant.
Show Solution
The derivative of the constant $\sqrt{a}$ is $0$.
$$ \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}y' = 0 $$
Move the first term to the other side.
$$ \frac{y'}{2\sqrt{y}} = -\frac{1}{2\sqrt{x}} $$
Multiply by $2\sqrt{y}$.
$$ \boxed{\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}} $$
Derivative of $b^2x^2 + a^2y^2 = a^2b^2$
Find $\dfrac{dy}{dx}$ if $b^2x^2 + a^2y^2 = a^2b^2$, where $a$ and $b$ are constants.
Show Solution
Constants such as $a^2$ and $b^2$ stay as multipliers.
$$ 2b^2x + 2a^2yy' = 0 $$
Solve for $y'$.
$$ 2a^2yy' = -2b^2x $$
$$ \boxed{\frac{dy}{dx} = -\frac{b^2x}{a^2y}} $$
Derivative of $(x-y)^3 = (x+y)^2$
Find $\dfrac{dy}{dx}$ if $(x-y)^3 = (x+y)^2$.
Show Solution
Use the chain rule on both sides. The derivative of $x-y$ is $1-y'$, and the derivative of $x+y$ is $1+y'$.
$$ 3(x-y)^2(1-y') = 2(x+y)(1+y') $$
Expand only enough to isolate $y'$.
$$ 3(x-y)^2 - 3(x-y)^2y' = 2(x+y) + 2(x+y)y' $$
Move non-$y'$ terms to the left and $y'$ terms to the right.
$$ 3(x-y)^2 - 2(x+y) = 3(x-y)^2y' + 2(x+y)y' $$
$$ 3(x-y)^2 - 2(x+y) = \left[3(x-y)^2 + 2(x+y)\right]y' $$
$$ \boxed{\frac{dy}{dx} = \frac{3(x-y)^2 - 2(x+y)}{3(x-y)^2 + 2(x+y)}} $$
Derivative of $y = 4(x^2 + y^2)$
Find $\dfrac{dy}{dx}$ if $y = 4(x^2 + y^2)$.
Show Solution
Differentiate both sides. On the right, multiply the derivative of the parentheses by $4$.
$$ y' = 4(2x + 2yy') $$
$$ y' = 8x + 8yy' $$
Collect the $y'$ terms.
$$ y' - 8yy' = 8x $$
$$ (1-8y)y' = 8x $$
$$ \boxed{\frac{dy}{dx} = \frac{8x}{1-8y}} $$
Derivative of $y^2 = \dfrac{3x+1}{2x-3}$
Find $\dfrac{dy}{dx}$ if $y^2 = \dfrac{3x+1}{2x-3}$.
Show Solution
Differentiate the left side by the chain rule and the right side by the quotient rule.
$$ 2yy' = \frac{(2x-3)(3) - (3x+1)(2)}{(2x-3)^2} $$
Simplify the numerator.
$$ (2x-3)(3) - (3x+1)(2) = 6x - 9 - 6x - 2 = -11 $$
$$ 2yy' = -\frac{11}{(2x-3)^2} $$
$$ \boxed{\frac{dy}{dx} = -\frac{11}{2y(2x-3)^2}} $$
Derivative of $y^2 - 3x + 2y = 0$
Find $\dfrac{dy}{dx}$ if $y^2 - 3x + 2y = 0$.
Show Solution
Differentiate each term with respect to $x$.
$$ 2yy' - 3 + 2y' = 0 $$
Collect the $y'$ terms.
$$ (2y+2)y' = 3 $$
$$ \boxed{\frac{dy}{dx} = \frac{3}{2(y+1)}} $$
Exercise 2.5 - Find $y''$
Find $y''$ in each equation. First find $y'$, then differentiate again and substitute $y'$ when needed.
Second Derivative of $xy = 32$
Find $y''$ if $xy = 32$.
Show Solution
Differentiate once using the product rule.
$$ xy' + y = 0 $$
$$ y' = -\frac{y}{x} $$
Differentiate $y' = -\dfrac{y}{x}$ using the quotient rule.
$$ y'' = -\frac{xy' - y}{x^2} = \frac{y - xy'}{x^2} $$
Substitute $y' = -\dfrac{y}{x}$.
$$ y'' = \frac{y - x\left(-\frac{y}{x}\right)}{x^2}
= \frac{2y}{x^2} $$
Since $xy = 32$, $y = \dfrac{32}{x}$.
$$ \boxed{y'' = \frac{64}{x^3}} $$
Second Derivative of $x^{2/3} + y^{2/3} = a^{2/3}$
Find $y''$ if $x^{2/3} + y^{2/3} = a^{2/3}$.
Show Solution
Differentiate once. The right side is constant, so its derivative is $0$.
$$ \frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}y' = 0 $$
Cancel $\dfrac{2}{3}$ and solve for $y'$.
$$ x^{-1/3} + y^{-1/3}y' = 0 $$
$$ y' = -\frac{y^{1/3}}{x^{1/3}} $$
Differentiate the equation $x^{-1/3} + y^{-1/3}y' = 0$ again. This avoids a messy quotient rule.
$$ -\frac{1}{3}x^{-4/3}
- \frac{1}{3}y^{-4/3}(y')^2
+ y^{-1/3}y'' = 0 $$
Solve for $y''$.
$$ y^{-1/3}y'' =
\frac{1}{3}x^{-4/3}
+ \frac{1}{3}y^{-4/3}(y')^2 $$
$$ y'' =
\frac{1}{3}y^{1/3}x^{-4/3}
+ \frac{1}{3}y^{-1}(y')^2 $$
Substitute $(y')^2 = \dfrac{y^{2/3}}{x^{2/3}}$.
$$ y'' =
\frac{1}{3}y^{1/3}x^{-4/3}
+ \frac{1}{3}y^{-1/3}x^{-2/3} $$
Factor and use $x^{2/3}+y^{2/3}=a^{2/3}$.
$$ y'' =
\frac{x^{2/3}+y^{2/3}}{3x^{4/3}y^{1/3}} $$
$$ \boxed{y'' = \frac{a^{2/3}}{3x^{4/3}y^{1/3}}} $$
Second Derivative of $y^2 - 16x = 0$
Find $y''$ if $y^2 - 16x = 0$.
Show Solution
Differentiate once.
$$ 2yy' - 16 = 0 $$
$$ y' = \frac{8}{y} $$
Differentiate $2yy' = 16$ again using the product rule.
$$ 2\left[(y')^2 + yy''\right] = 0 $$
$$ yy'' = -(y')^2 $$
Substitute $y' = \dfrac{8}{y}$.
$$ y'' = -\frac{(y')^2}{y}
= -\frac{64/y^2}{y} $$
$$ \boxed{y'' = -\frac{64}{y^3}} $$
Second Derivative of $x^2 - 2xy + 3y^2 = 4$
Find $y''$ if $x^2 - 2xy + 3y^2 = 4$.
Show Solution
Differentiate once.
$$ 2x - 2(xy' + y) + 6yy' = 0 $$
$$ 2x - 2y + (-2x + 6y)y' = 0 $$
$$ y' = \frac{y-x}{3y-x} $$
Differentiate $y' = \dfrac{y-x}{3y-x}$ using the quotient rule.
$$ y'' =
\frac{(y'-1)(3y-x) - (y-x)(3y'-1)}{(3y-x)^2} $$
Use $y' = \dfrac{y-x}{3y-x}$ to simplify the two small factors.
$$ y' - 1 =
\frac{y-x-(3y-x)}{3y-x}
= -\frac{2y}{3y-x} $$
$$ 3y' - 1 =
\frac{3(y-x)-(3y-x)}{3y-x}
= -\frac{2x}{3y-x} $$
Substitute these into the quotient-rule expression.
$$ y'' =
\frac{-2y + \frac{2x(y-x)}{3y-x}}{(3y-x)^2} $$
$$ y'' =
\frac{-2y(3y-x)+2x(y-x)}{(3y-x)^3} $$
$$ y'' =
\frac{-2x^2+4xy-6y^2}{(3y-x)^3}
= -\frac{2(x^2-2xy+3y^2)}{(3y-x)^3} $$
Since $x^2 - 2xy + 3y^2 = 4$, replace the grouped expression by $4$.
$$ \boxed{y'' = -\frac{8}{(3y-x)^3}} $$
Second Derivative of $4x^2 + 9y^2 = 36$
Find $y''$ if $4x^2 + 9y^2 = 36$.
Show Solution
Differentiate once.
$$ 8x + 18yy' = 0 $$
$$ y' = -\frac{4x}{9y} $$
Differentiate $8x + 18yy' = 0$ again.
$$ 8 + 18\left[(y')^2 + yy''\right] = 0 $$
$$ y'' = -\frac{8 + 18(y')^2}{18y} $$
Substitute $y' = -\dfrac{4x}{9y}$.
$$ y'' =
-\frac{8 + 18\left(\frac{16x^2}{81y^2}\right)}{18y} $$
$$ y'' =
-\frac{36y^2 + 16x^2}{81y^3}
= -\frac{4(9y^2+4x^2)}{81y^3} $$
Since $4x^2 + 9y^2 = 36$, the grouped expression $9y^2+4x^2$ is $36$.
$$ \boxed{y'' = -\frac{16}{9y^3}} $$
Exercise 2.5 - Slopes at Given Points
Find the slope of the curve at the given point. The slope is the value of $\dfrac{dy}{dx}$ after substituting the point.
Slope of $2x^3 + 2y^3 = 9xy$ at $(2,1)$
Find the slope of $2x^3 + 2y^3 = 9xy$ at $(2,1)$.
Show Solution
Differentiate implicitly. Use the product rule on $xy$.
$$ 6x^2 + 6y^2y' = 9(xy' + y) $$
Collect the $y'$ terms.
$$ 6y^2y' - 9xy' = 9y - 6x^2 $$
$$ (6y^2 - 9x)y' = 9y - 6x^2 $$
$$ y' = \frac{9y - 6x^2}{6y^2 - 9x} $$
Substitute $(x,y)=(2,1)$.
$$ m = \frac{9(1) - 6(2)^2}{6(1)^2 - 9(2)}
= \frac{9 - 24}{6 - 18}
= \frac{-15}{-12} $$
$$ \boxed{m = \frac{5}{4}} $$
Slope of $y^3 = x^2 - 1$ at $(3,2)$
Find the slope of $y^3 = x^2 - 1$ at $(3,2)$.
Show Solution
Differentiate both sides.
$$ 3y^2y' = 2x $$
$$ y' = \frac{2x}{3y^2} $$
Substitute $(x,y)=(3,2)$.
$$ m = \frac{2(3)}{3(2)^2}
= \frac{6}{12} $$
$$ \boxed{m = \frac{1}{2}} $$
Slope of $x^2 + 4\sqrt{xy} + y^2 = 25$ at $(4,1)$
Find the slope of $x^2 + 4\sqrt{xy} + y^2 = 25$ at $(4,1)$.
Show Solution
The photo shows the point beginning with $(4,\ldots)$. The visible curve is satisfied by $(4,1)$, so this solution uses $(4,1)$.
Differentiate the equation. For $\sqrt{xy}$, write it as $(xy)^{1/2}$ and use the chain rule plus product rule.
$$ 2x + 4\left[\frac{1}{2\sqrt{xy}}(xy' + y)\right] + 2yy' = 0 $$
$$ 2x + \frac{2(xy' + y)}{\sqrt{xy}} + 2yy' = 0 $$
Divide by $2$ to simplify.
$$ x + \frac{xy' + y}{\sqrt{xy}} + yy' = 0 $$
Collect the $y'$ terms.
$$ \left(\frac{x}{\sqrt{xy}} + y\right)y'
= -x - \frac{y}{\sqrt{xy}} $$
$$ y' =
-\frac{x + \frac{y}{\sqrt{xy}}}{\frac{x}{\sqrt{xy}} + y} $$
At $(4,1)$, $\sqrt{xy}=\sqrt{4(1)}=2$.
$$ m =
-\frac{4 + \frac{1}{2}}{\frac{4}{2}+1}
= -\frac{\frac{9}{2}}{3} $$
$$ \boxed{m = -\frac{3}{2}} $$
Slope of Tangent via Implicit Differentiation — CE Board
Find the slope of the tangent to the curve $x^2 + y^2 = 25$ at the point $(3,\,4)$.
Show Solution
Differentiate both sides implicitly with respect to $x$:
$2x + 2y\dfrac{dy}{dx} = 0$
Solve for $\dfrac{dy}{dx}$:
$\dfrac{dy}{dx} = -\dfrac{x}{y}$
Substitute $(3,\,4)$:
$\dfrac{dy}{dx}\bigg|_{(3,4)} = -\dfrac{3}{4}$
The slope of the tangent is $-\dfrac{3}{4}$.
Implicit Differentiation with Mixed Terms
Find $\dfrac{dy}{dx}$ given $x^2 + xy + y^2 = 7$.
Show Solution
Differentiate every term with respect to $x$ (apply product rule to $xy$):
$2x + \left(y + x\dfrac{dy}{dx}\right) + 2y\dfrac{dy}{dx} = 0$
Collect $\dfrac{dy}{dx}$ terms on one side:
$(x + 2y)\dfrac{dy}{dx} = -(2x + y)$
$\boxed{\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}}$
Implicit Differentiation — Trigonometric
Find $\dfrac{dy}{dx}$ if $\sin(x + y) = 2x$.
Show Solution
Differentiate both sides using the chain rule:
$\cos(x+y)\cdot\left(1 + \dfrac{dy}{dx}\right) = 2$
Solve for $\dfrac{dy}{dx}$:
$1 + \dfrac{dy}{dx} = \dfrac{2}{\cos(x+y)}$
$\boxed{\dfrac{dy}{dx} = \dfrac{2}{\cos(x+y)} - 1 = 2\sec(x+y) - 1}$
Equation of Tangent Line — CE Board Style
Find the equation of the tangent to the circle $x^2 + y^2 - 6x - 4y + 8 = 0$ at the point $(1,\,1)$.
Show Solution
Verify the point: $1 - 6 + 1 - 4 + 8 = 0$ ✓
Differentiate implicitly:
$2x + 2y\dfrac{dy}{dx} - 6 - 4\dfrac{dy}{dx} = 0$
$\dfrac{dy}{dx} = \dfrac{6 - 2x}{2y - 4} = \dfrac{3 - x}{y - 2}$
At $(1,\,1)$:
$m = \dfrac{3-1}{1-2} = \dfrac{2}{-1} = -2$
Using point-slope form:
$y - 1 = -2(x - 1)$
$\boxed{y = -2x + 3}$
Second-Order Implicit Derivative — CE Board
Given the curve $x^2 + y^2 = 25$, find the second derivative $\dfrac{d^2y}{dx^2}$.
Show Solution
Step 1: Find the first derivative.
$2x + 2y\,y' = 0 \implies y' = -\dfrac{x}{y}$
Step 2: Differentiate $y' = -\dfrac{x}{y}$ using the quotient rule.
$y'' = -\dfrac{(1)(y) - x\,y'}{y^2}$
Substitute $y' = -x/y$:
$y'' = -\dfrac{y - x\!\left(-\dfrac{x}{y}\right)}{y^2} = -\dfrac{y + \dfrac{x^2}{y}}{y^2} = -\dfrac{y^2 + x^2}{y^3}$
Since $x^2 + y^2 = 25$:
$\boxed{y'' = -\dfrac{25}{y^3}}$
Exam Generator Problems
Additional board-style practice items for this topic.
Question Bank: q436
MSTE - Differential Calculus / Implicit Differentiation / Engr. Janclyde Espinosa (Clidez)
Find dy/dx'(x2 -4y2 =4.
Answer:
x/4y
-x/4y
4y/x
-4x/y
Show Solution
Differentiate implicitly:
Question Bank: t780
MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6
Find the obtuse angle between the tangents to the circle x^2 + y^2 - 2x + 4y - 20 = 0 at points (-2, -6) and (4, -6).
36.87°
52.47°
73.74°
106.26°
Show Solution
The circle has center $(1, -2)$. Each tangent is perpendicular to its radius, so the angle between tangents equals the angle between radii.
Question Bank: t790
MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6
Find the acute angle between the tangents to the circle x^2 + y^2 - 2x + 4y - 20 = 0 at points (- 2, -6) and (4, -6).
73.74°
36.87°
52.47°
63.58°
Show Solution
The circle has center $(1, -2)$. Each tangent is perpendicular to its radius, so the angle between tangents equals the angle between radii.
Question Bank: t824
MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6
A stone contractor has his quarry 10 km from P, the nearest point to a straight railway. The railroad company agrees to haul his stone to S, 30 km along the straight track from P for P30 per ton per km. The cost of hauling by truck P100 per ton per km.
How far from P should the contractor drop his stone in order that the cost of hauling is minimum?
2.785 km
3.145 km
3.525 km
5.325 km
What is the total hauling distance from the quarry to S for minimum hauling cost?
37.34 km
38.65 km
35.14 km
34.21 km
What is the minimum cost of hauling his stone to S?
P1,543 per ton
P1,655 per ton
P1,854 per ton
P2,321 per ton
Show Solution
Part 1.
Let the drop point be $x$ km from $P$. Cost $C = 100\sqrt{100 + x^2} + 30(30 - x)$.
$\frac{dC}{dx} = \frac{100x}{\sqrt{100 + x^2}} - 30 = 0 \Rightarrow 100x = 30\sqrt{100 + x^2}$
$10000x^2 = 900(100 + x^2) \Rightarrow 9100x^2 = 90000$
$\boxed{x = 3.145 \text{ km}}$
Part 2.
Total distance = truck haul + rail haul:
$\sqrt{100 + 3.145^2} + (30 - 3.145) = 10.48 + 26.86$
$\boxed{37.34 \text{ km}}$
Part 3.
Minimum cost using $x = 3.145$:
$C = 100\sqrt{100 + 3.145^2} + 30(30 - 3.145)$
$= 100(10.48) + 30(26.86) = 1048 + 806$
$\boxed{\text{P}1{,}854 \text{ per ton}}$
Question Bank: t861
MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6
A steel ball is dropped from the surface of a body of water. The acceleration of the ball at any instant is defined by the equation $a = 0.90g - c v^2$, where g is the acceleration due to gravity (g = 9.81 m/s^2), c = 3.02 m⁻¹ and v is the velocity of the ball in m/s.
What is the velocity of the ball after 0.25 second?
1.98
1.12
1.47
2.13
What is the distance travelled by the ball after 0.25 second?
0.54 m
0.22 m
0.89 m
0.43 m the next batch of questions (106–120)!
What is the acceleration of the ball after 0.25 second?
1.23 m/s^2
2.31 m/s^2
3.21 m/s^2
1.32 m/s^2
Show Solution
Part 1.
With $\frac{dv}{dt} = 0.9g - cv^2 = 8.829 - 3.02v^2$, the solution is $v = v_t\tanh(kt)$, where $v_t = \sqrt{\frac{0.9g}{c}} = 1.710$ m/s and $k = \sqrt{0.9g\cdot c} = 5.164$.
At $t = 0.25$ s:
$v = 1.710\tanh(0.25\times5.164) = 1.710\tanh(1.291)$
$\boxed{v = 1.47 \text{ m/s}}$
Part 2.
Distance is $x = \int v\,dt = \frac{v_t}{k}\ln(\cosh kt)$.
At $t = 0.25$ s:
$x = \frac{1.710}{5.164}\ln\!\big(\cosh 1.291\big) = 0.331\ln(1.956)$
$\boxed{x = 0.22 \text{ m}}$
Part 3.
Acceleration at $t = 0.25$ s uses $v = 1.47$ m/s:
$a = 0.9g - cv^2 = 8.829 - 3.02(1.47)^2$
$= 8.829 - 6.52$
$\boxed{a = 2.31 \text{ m/s}^2}$
Question Bank: t890
MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6
A plane area is bounded in the first quadrant by the curves y^2 = -16(x - 4), the line 8x - 15y = 0, and the x- axis.
What is the slope of the line?
-15/8
15/8
- 8/15
8/15 as identical values in the source text. It is mapped exactly as printed.) next batch (136–150)!
What is the area bounded by the curves?
7.25
4.08
9.87
10.54
What is the moment of inertia of this area about the x-axis?
2.77
4.58
9.65
6.21
Show Solution
Write the line in slope-intercept form: $8x - 15y = 0 \Rightarrow y = \frac{8}{15}x$.
Question Bank: t1467
MSTE - Differential Calculus / Implicit Differentiation / BEMz
Find dy/dx by implicit differentiation at the point (3,4) when $x^{2}+y^{2}=25$.
-3/4
¾
2/3
-2/3
Solution pending in psadquestions/t1467.json.
Question Bank: t1468
MSTE - Differential Calculus / Implicit Differentiation / BEMz
Find dy/dx by implicit differentiation at point (0,0) if $(x^{3})(y^{3})-y=x$.
-1
-2
2
1
Solution pending in psadquestions/t1468.json.
Question Bank: t1469
MSTE - Differential Calculus / Implicit Differentiation / BEMz
Find dy/dx by implicit differentiation at point (0,-2) if $x^{3}-xy+y^{2}=4$.
½
-2
-2/3
¾
Solution pending in psadquestions/t1469.json.
