How to find rate of change - AP Calculus AB

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Question

A cube is growing in size. What is the rate of growth of the cube's surface area if its sides have a length of 8 and a rate of growth of 33?

Answer

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

The rates of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=12s\frac{ds}{dt}$

Now, with the rate equation known, we can solve for the rate of change of the surface area with what we know about the cube, namely that its sides have a length of 8 and a rate of growth of 33:

$\frac{dA}{dt}=12(8)(33)=3168$

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Question

A cube is growing in size. What is the rate of growth of the cube's surface area if its sides have a length of 7 and a rate of growth of 34?

Answer

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=12s\frac{ds}{dt}$

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 7 and a rate of growth of 34:

$\frac{dA}{dt}=12(7)(34)=2856$

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Question

A cube is growing in size. What is the rate of growth of the cube's surface area if its sides have a length of 6 and a rate of growth of 35?

Answer

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=12s\frac{ds}{dt}$

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 6 and a rate of growth of 35:

$\frac{dA}{dt}=12(6)(35)=2520$

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Question

A cube is growing in size. What is the rate of growth of the cube's surface area if its sides have a length of 5 and a rate of growth of 36?

Answer

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=12s\frac{ds}{dt}$

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 5 and a rate of growth of 36:

$\frac{dA}{dt}=12(5)(36)=2160$

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Question

A cube is growing in size. What is the rate of growth of the cube's surface area if its sides have a length of 4 and a rate of growth of 37?

Answer

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=12s\frac{ds}{dt}$

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 4 and a rate of growth of 37:

$\frac{dA}{dt}=12(4)(37)=1776$

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Question

A cube is growing in size. What is the rate of growth of the cube's surface area if its sides have a length of 3 and a rate of growth of 38?

Answer

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=12s\frac{ds}{dt}$

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 3 and a rate of growth of 38:

$\frac{dA}{dt}=12(3)(38)=1368$

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Question

A cube is growing in size. What is the rate of growth of the cube's surface area if its sides have a length of 2 and a rate of growth of 39?

Answer

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=12s\frac{ds}{dt}$

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 2 and a rate of growth of 39:

$\frac{dA}{dt}=12(2)(39)=936$

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Question

A cube is growing in size. What is the rate of growth of the cube's surface area if its sides have a length of 1 and a rate of growth of 40?

Answer

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=12s\frac{ds}{dt}$

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 1 and a rate of growth of 40:

$\frac{dA}{dt}=12(1)(40)=480$

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Question

Find the rate of change of from $x:(0, \pi)$

Answer

To do rate of change , remember it is equivalent to finding slope.

$m=\frac{f(\pi)-f(0)}{\pi-0}=\frac{sec(2\pi)-sec(0)}{\pi-0} =\frac{1-1}{\pi-0} = 0$

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Question

The position of a car is defined by the equation $r(t)=5t^{2}-7t+6$ . What is the average velocity of the car between and ?

Answer

The average velocity of an object between and is given by the equation

$v_{average}=\frac{\Delta y}{\Delta x}=\frac{r(b)-r(a)}{b-a}$

In this problem,

$r(4)=5(4)^{2}-7(4)+6=80-28+6=58$

$r(2)=5(2)^{2}-7(2)+6=20-14+6=0$

$v_{average}=\frac{r(4)-r(2)}{4-2}=\frac{58-0}{4-2}=29$

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Question

The position of a car is defined by the equation $r(t)=5t^{2}-7t+6$ . What is the instantaneous velocity of the car at ?

Answer

The instantaneous velocity of the car is the first derivative of the position at a given point.

In this problem,

$r(t)=5t^{2}-7t+6$

$v_{ins}=r'(t)=10t-7$

$v_{ins}=r'(4)=10*4-7=40-7=33$

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Question

Find the slope of the line tangent to the curve of the multivariable function f(x,y) at

the point $(\frac{\sqrt{\pi }}{2},\frac{\sqrt{\pi }}{2})$ .

Answer

To find the slope of the tangent line at the specified point, we must first verify that the specified point actually exists on the curve. We check that

$1=sin(x^2+y^2)=sin((\frac{\sqrt{\pi }}{2})^2+(\frac{\sqrt{\pi }}{2})^2)=sin(\frac{\pi }{2})=1$

Since the verification checks out, the problem has a solution and we can continue with Implicit Differentiation.

Recall that for Implicit differentiation, if we have a function in terms of y, we have that it's derivative with respect to x is

$\frac{d}{dx}[f(y)]=\frac{df}{dy}\frac{dy}{dx}$

Applying this to the given function, we have that

$\frac{d}{dx}[1=sin(x^2+y^2)]$

We must also utilize the Chain Rule to obtain the derivative; we get that

$0=cos(x^2+y^2)[2x+2y\frac{dy}{dx}]$

Algebraically, we divide the cosine term to begin isolating dy/dx. We then get that

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

$-2x=2y\frac{dy}{dx}$

$\frac{dy}{dx}=-\frac{x}{y}$

To obtain the slope of the tangent line, we substitute the specified point (x,y) for x and y respectively.

$\frac{dy}{dx}=-\frac{\frac{\sqrt{\pi }}{2}}{\frac{\sqrt{\pi }}{2}}=-1$

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Question

Find $\frac{dy}{dx}$ for $f(x)= \frac{ln(x)}{x^8}$ .

Answer

To solve this problem, we can use either the quotient rule or the product rule. For this solution, we will use the product rule.

The product rule states that $\frac{d(u*v)}{dx}=u'v+v'u$ .

In this case, let $u=x^{-8}\Rightarrow u'=-8x^{-9}$ and $v=ln(x)\Rightarrow v'=\frac{1}{x}$ .

Putting both of these together, we get

$\frac{d(x^{-8}ln(x))}{dx}=-8x^{-9}ln(x)+\frac{1}{x}x^{-8}=\frac{1}{x^{9}}+\frac{-8ln(x)}{x^9}=\frac{1-8ln(x)}{x^9}$ .

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Question

A leaky trough is ten feet long with isosceles triangle cross sections. The cross sections have a base of two feet and a height of two feet six inches. The trough is being filled with water at one cubic foot per minute. However, it is also leaking at a rate of two cubic feet per minute.

When the depth of the water is one foot five inches, how fast is the water level falling?

Answer

You know the net volume is decreasing at a rate of -1 ft/min by adding the rates 1 (being added) and -2(leaking from the trough). However, the question asks what the rate of change of the height is. The equation V=1/2blh (because the cross sections are triangles; the trough is a prism) relates height to volume.

The length (l) is a constant 10 feet, and the base needs to be written in terms of something we know the rate of change. Because the cross sections are triangles, the sides are proportional.

Therefore, $\frac{b}{2}=\frac{h}{2.5}$ and b=0.8h.

After plugging the known values into the volume equation,

$V=\frac{1}{2}100.8h*h$ or .

Then differentiate both sides to relate the rates of change.

$\frac{dV}{dt}=8h\frac{dh}{dt}$ .

Finally, plug in the known values for the rate of change of volume(dV/dt) -1ft/min and the instantaneous height (1 ft 5 in = 17/12 ft).

$\ -1=8\left(\frac{17}{12} \right )\frac{dh}{dt}\ \-1=\frac{34}{3}\frac{dh}{dt} \ \=-\frac{3}{34}=\frac{dh}{dt}$

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Question

Determine the average rate of change of the function from the interval $\left[\frac{\pi}{2},\pi\right]$ .

Answer

Write the formula to determine average rate of change.

$\frac{\Delta f}{\Delta x}= \frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$

Substitute the values and solve for the average rate of change.

$\frac{-cos(\pi)-(-cos(\frac{\pi}{2}))}{\pi-\frac{\pi}{2}} = \frac{-(-1)+0}{\frac{\pi}{2}}=\frac{1}{\frac{\pi}{2}}=\frac{2}{\pi}$

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Question

Find $\frac{dV}{dt}$ if the radius of a spherical balloon is increasing at a rate of per second.

Answer

The volume function, in terms of a radius , is given as

$V(r)=\frac{4}{3}\pi r^3$ .

The change in volume over the change in time, or

$\frac{dV}{dt}$ is given as

$\frac{dV}{dt} = \frac{d}{dt}[V(r)]$

and by implicit differentiation, the chain rule, and the power rule,

$\frac{d}{dt}[V(r)]=\frac{d}{dt}[\frac{4}{3}\pi r^3]=4\pi r^2 \frac{dr}{dt}$ .

Setting $\frac{dr}{dt}=2$ we get

$\frac{d}{dt}[V(r)]= 4\pi r^2(2)=8\pi r^2$ .

As such,

$\frac{dV}{dt} =8\pi r^2$ .

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Question

Find the rate of change of a function from $x_{1}=3$ to $x_{2}=6$ .

Answer

Write the formula for the average rate of change from the interval $[{x_{1},x_{2}}]$ .

$\frac{f(x_2)-f(x_{1})}{x_2-x_1}$

Solve for and .

Substitute the known values into the formula and solve.

$\frac{f(x_2)-f(x_{1})}{x_2-x_1}=\frac{3-(-3)}{6-3}=\frac{6}{3}=2$

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Question

Suppose the rate of a square is increasing at a constant rate of meters per second. Find the area's rate of change in terms of the square's perimeter.

Answer

Since the question is asking for the rate of change in terms of the perimeter, write the formula for the perimeter of the square and differentiate it with the respect to time.

$\frac{dP}{dt}= 4\frac{ds}{dt}$

The question asks in terms of the perimeter. Isolate the term by dividing four on both sides.

$s=\frac{P}{4}$

Write the given rate in mathematical terms and substitute this value into $\frac{dP}{dt}$ .

$\frac{ds}{dt}=2$

$\frac{dP}{dt}= 4\frac{ds}{dt}=4(2)=8$

Write the area of the square and substitute the side.

$A=s^2=\left(\frac{P}{4}\right)^2=\frac{P^2}{16}$

Since the area is changing with time, take the derivative of the area with respect to time.

$\frac{dA}{dt}=\frac{P}{8} \cdot \frac{dP}{dT}$

Substitute the value of $\frac{dP}{dt}$ .

$\frac{dA}{dt}=\frac{P}{8} \cdot 8=P$

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Question

You are looking at a balloon that is away. If the height of the balloon is increasing at a rate of , at what rate is the angle of inclination of your position to the balloon increasing after seconds?

Answer

Using right triangles we know that

$\tan(\theta(t))= \frac{h(t)}{300} \right$ .

Solving for $\theta(t)$ we get

$\theta(t)=\arctan\left ( \frac{h(t)}{300} \right )$ .

Taking the derivative, we need to remember to apply the chain rule to since the height depends on time,

$\frac{d\theta}{dt}=\frac{1}{1+\left(\frac{h}{300} \right )^2}\frac{1}{300}\frac{dh}{dt}$ .

We are asked to find $\frac{d\theta}{dt}|_{t=5}$ . We are given $\frac{dh}{dt}=50$ and since $\frac{dh}{dt}$ is constant, we know that the height of the balloon is given by $h(t)=\frac{dh}{dt}t$ .

Therefore, at we know that the height of the balloon is $h(5)=5\cdot 50=250$ .

Plugging these numbers into $\frac{d\theta}{dt}$ we find

$\frac{d\theta}{dt}|_{t=5} =\frac{1}{1+\left(\frac{250}{300} \right )^2}\frac{1}{300}50=0.0098$ radians.

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Question

Boat leaves a port at noon traveling . At the same time, boat leaves the port traveling east at . At what rate is the distance between the two boats changing at ?

Answer

This scenario describes a right triangle where the hypotenuse is the distance between the two boats. Let denote the distance boat is from the port, denote the distance boat is from the port, denote the distance between the two boats, and denote the time since they left the port. Applying the Pythagorean Theorem we have,

.

Implicitly differentiating this equation we get

$2D(t)\frac{dD}{dt}=2x(t)\frac{dx}{dt}+2y(t)\frac{dy}{dt}$ .

We need to find $\frac{dD}{dt}$ when .

We are given

$\frac{dx}{dt}=30, \frac{dy}{dt}=20$ which tells us

$x(3)=3\cdot 30=90, y(3)=3\cdot 20=60,$

$D(3)=\sqrt{90^2+60^2}=30\sqrt{13}$ .

Plugging this in we have

$2\cdot 30\sqrt{13}\frac{dD}{dt}|_{t=3}=2\cdot90\cdot 30+2\cdot60\cdot20$ .

Solving we get

$\frac{dD}{dt}|_{t=3}=10\sqrt{13}$ .

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