Rate of Change - AP Calculus AB

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Question

A sphere is fixed inside of a cube, such that it is completely snug. If the sides of the cube, which have length , begin to grow at a rate of , what is the rate of growth of the volume of the sphere?

Answer

The volume of sphere, in terms of its radius, is defined as

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

However, in the case of the problem, we're given the lengths of the sides of a cube in which the sphere fits. Since the outside of the sphere is touching the sides walls of the cube, the length of the cube is the diameter of the sphere:

$r=\frac{s}{2}$

$r=\frac{6}{2}=3$

Furthermore, the rate of growth of a sphere's radius will be half the rate of growth of it's diameter

$\frac{dr}{dt}=\frac{1}{2}\frac{ds}{dt}$

$\frac{dr}{dt}=\frac{1}{2}(0.4)=0.2$

Now returning to the volume equation

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

The rate of growth can be found by taking the derivative with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

$\frac{dV}{dt}=4\pi (6^2)(0.2)$

$\frac{dV}{dt}=28.8\pi$

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Question

The rate of change of the radius of a sphere is $9e^{-t}$ . If the sphere has an initial radius of , what is the rate of change of the sphere's surface area at time ?

Answer

To say that the rate of change of the radius of a sphere is $9e^{-t}$ means

$\frac{dr(t)}{dt}=9e^{-t}$

The equation for the length of the radius can be found by integrating this equation with respect to time:

$\int e^{nt}dt=\frac{e^{nt}}{n}+C$

$r(t)=-9e^{-t}+C$

The constant of integration can be found by utilizing the initial condition:

$r(0)=-9e^{0}+C=6$

$r(t)=15-9e^{-t}$

The surface area of a sphere is given by the equation

$A=4\pi r^2$

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

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$

$\frac{dA(t)}{dt}=8\pi (15-9e^{-t}) (9e^{-t})$

$\frac{dA(ln(3))}{dt}=8\pi (15-9e^{-ln(3)}) (9e^{-ln(3)})$

$\frac{dA(ln(3))}{dt}=288\pi$

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Question

We can interperet a derrivative as $\frac{dy}{dx}= \lim_{\Delta \rightarrow 0} \frac{\Delta y}{\Delta x}$ (i.e. the slope of the secant line cutting the function as the change in x and y approaches zero) but these so-called "differentials" ( $\Delta x$ and $\Delta y$ ) can be a good tool to use for aproximations. If we suppose that $\frac{dy}{dx}=f'(x)$ , or equivalently . If we suppose a change in x (have a concrete value for $\Delta x$ ) we can find the change in with the afore mentioned relation.

Let $f(x)= \sin( \ln x)$ . Find $f'(e^{\frac{\pi}{3}})$ let and $\Delta x =0.1$ . Find $\Delta y$ under such conditions.

Answer

We find the derivative of the function:

$f(x)= \sin( \ln x)$

$\frac{dy}{dx}=\frac{\cos(\ln x)}{x}$

Evaluating at $x=e^{\frac{\pi}{3}}$

$\frac{dy}{dx}|_e^{\frac{\pi}{3}} =\frac{\cos(\ln (e^\frac{\pi}{3}))}{(e^\frac{\pi}{3})}$

$\frac{dy}{dx}=\frac{1}{2}{e^{-\frac{\pi}{3}}} \approx 0.1754$

$dy=\frac{1}{2}{e^{-\frac{\pi}{3}}}dx$

Letting

$dy=\frac{1}{2}{e^{-\frac{\pi}{3}}}*.1 \approx 0.01754$

Which is our answer.

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Question

A regular tetrahedron is increaseing in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its height when its sides have length $\sqrt[3]{6}$ ?

Answer

To solve this problem, define a regular tetrahedron's dimensions, its volume and height in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$h=\frac{\sqrt{6}}{3}s$

Rates of change can then be found by taking the derivative of each property with respect to time:

$\frac{dV}{dt}=\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}$

$\frac{dh}{dt}=\frac{\sqrt{6}}{3}\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(\phi)\frac{\sqrt{6}}{3}\frac{ds}{dt}$

$\phi=\frac{\sqrt{3}s^2}{4}$

$\phi=\frac{\sqrt{3}(\sqrt[3]{6})^2}{4}=\frac{3\sqrt[6]{3}}{2\sqrt[3]{2}}$

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Question

Find the rate of change of a function from to .

Answer

We can solve by utilizing the formula for the average rate of change: $\frac{f(x_2)-f(x_1)}{x_2-x_1}$ .

Solving for f(x) at our given points:

Plugging our values into the average rate of change formula, we get:

$\frac{(26-21)}{(3-2)}=\frac{5}{1}=5$ .

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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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Question

Determine the point on the function that is not changing:

Answer

In order to determine where the function is not changing, it is necessary to take the derivative and set the slope equal to zero. This will provide information on where the curve is not changing. Once we find the x value that gives the derivative a slope of zero, we can substitute the x-value back into the original function to obtain the point.

Substitute this value back to the original equation to solve for .

The point where the function is not changing is .

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Question

A rectangle has a length of four feet and a width of six feet. If the width of the rectangle increases at a rate of $2\frac{ft}{sec}$ , how fast is the area of the rectangle increasing?

Answer

In this problem we are given the length and width of a rectangle as well as the rate at which the width is increasing. We are asked to find the rate of change of the area of a rectangle. The equation for finding the area of a rectangle is given as

$A=l\cdot w$ .

By taking the derivative of this equation with respect to time, we can find how the area changes with respect to time. To take the derivative of an equation with two variables, we must use the product rule,

$(uv)^{'}=u^{'}v+uv^{'}$ .

Applying the product rule to the equation we obtain

$\frac{dA}{dt}=l\frac{dw}{dt}+w\frac{dl}{dt}$ .

Because the width of the rectangle is increases at a rate of $2\frac{ft}{s}$ , $\frac{dw}{dt}=2$ .

Since the length of the rectangle does not change with respect to time, $\frac{dl}{dt}=0$ .

and are given to us as 4 feet and 6 feet respectively $\frac{dA}{dt}=8$ .

Therefore the area of this rectangle changes at a rate of when the width of the rectangle is increasing by $2\frac{ft}{s}$ .

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Question

At what time does the function $f(t)=t^2-e^{2t}$ have a slope of ? Round to the nearest hundredth.

Answer

First, we want the slope, so we have to take a derivative of . We will need to use the power rule which is,

$y=x^n \rightarrow \frac{d}{dx}=nx^{n-1}$ on the first term. We need to also recall that the derivative of is $e^u\cdot \frac{du}{dx}$ .

Applying these rules we get the following derivative.

$\frac{\mathrm{d} }{\mathrm{d} t}(t^2-e^{2t})=2t-2e^{2t}$

We're looking for the time the slope is , so we have to set the derivative (which gives you slope) equal to .

$2t-2e^{2t}=-10 \therefore t-e^{2t}=-5$ .

At this point you can use a graphing calculator to graph the function $t-e^{2t}$ , and trace the graph to find the x value that results in a y value of . The positive solution rounded to the nearest hundredth is .

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Question

A spherical balloon is inflating at rate of 0.1 cubic meters per second. Find the rate of change of the surface area of the balloon (in square meters per second) when the balloon has a radius of 1 meter.

Answer

The volume of a sphere is given by

$V=\frac{4\pi }{3}r^3$ , where r is the radius of the sphere.

The surface area of a sphere is given by $A=4\pi r^2$ (r is the radius of the sphere). The rate of change of the volume of the sphere is found by taking the first derivative of the function for volume:

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$ , where $\frac{dr}{dt}$ is the change of the radius with respect to time.

This derivative was found by using the power rule

$\frac{d}{dt}x^n=nx^{n-1}\frac{dx}{dt}$ .

The rate of change of the surface area of the sphere is found the same way, instead using the function for surface area:

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$ .

The balloon's volume is increasing at a constant rate of 0.1 cubic meters per second when the radius of the balloon is 1 meter, thus plugging in these values into the $\frac{dV}{dt}$ equation gives us the following:

$0.1 \frac{m^3}{sec}=4\pi (1^2)\frac{dr}{dt}$ . Solving for $\frac{dr}{dt}$ , we get $\frac{dr}{dt}=\frac{1}{40\pi }$ meters per second. Since we are solving for the rate of change of surface area of the balloon, we plug this value into its equation along with the radius we want (1 m). Solving for the rate of change of the surface area we get:

$\frac{dA}{dt}=8\pi (1)\left(\frac{1}{40\pi }\right)=\frac{1}{5}$ square meters per second.

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Question

For the function , what is the average rate of change from to ?

Answer

Write the formula for average rate of change.

$\frac{f(x_f)-f(x_i)}{x_f-x_i}$

Determine the values of and .

Substitute the known values.

$\frac{f(x_f)-f(x_i)}{x_f-x_i}=\frac{10-10}{3-1}=0$

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Question

Find the rate of change of a function from to .

Answer

We can solve by utilizing the formula for the average rate of change: $\frac{f(x_2)-f(x_1) }{x_2-x_1}$ . Solving for at our given points:

Plugging our values into the average rate of change formula, we get:

$\frac{(16-13)}{(3-2)}=\frac{3}{1}=3$

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Question

Find the rate of change of a function from to .

Answer

We can solve by utilizing the formula for the average rate of change: $\frac{f(x_2)-f(x_1)}{x_2-x_1}.$ Solving for at our given points:

Plugging our values into the average rate of change formula, we get:

$\frac{(39-29)}{(6-4)}=\frac{10}{2}=5$

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