Rate - AP Calculus AB

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Question

The initial position of a particle is 0. If its velocity is described by v(t) = 4t + 5, what is its position at the time when the velocity is equal to 45?

Answer

The position s(t) is equal to ∫v(t) dt = ∫ 4t + 5 dt = 2t2 + 5t + C

Now, since we know that the initial position of the particle (at t = 0) is 0, we know C is 0. Therefore, s(t) = 2t2 + 5t + C

To find the time t when at which the velocity is 45, set v(t) equal to 45. 45 = 4t + 5 → 40 = 4t → t = 10

The position of the particle is s(10) = 2 * 102 + 50 = 200 + 50 = 250

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Question

The position of a particle is defined by s(t) = 4t3 – 3t2 + 2t. At what time (to the nearest hundreth) is its velocity equal to 384?

Answer

The position of a particle is defined by s(t) = 4t3 – 3t2 + 2t. At what time (to the nearest hundreth) is its velocity equal to 384?

Start by finding the velocity function:

v(t) = s'(t) = 12t2 – 6t + 2

To find the time, set v(t) equal to 384: 384 = 12t2 – 6t + 2

To solve, set equal the equation equal to 0: 12t2 – 6t – 382 = 0

Use the quadratic formula:

(–(–6) ± √(36 – 412(–382)))/(2 * 12)

= (6 ± √(36 + 18336))/24 = (6 ± √(18372))/24 = (6 ± 2√(4593))/24

= (3 ± √(4593))/12

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Question

The initial position of a particle is 44.5. If its velocity is described by v(t) = 3t + 12, what is its position at the time when the velocity is equal to 8391?

Answer

The initial position of a particle is 44.5. If its velocity is described by v(t) = 3t + 12, what is its position (to the nearest hundreth) at the time when the velocity is equal to 8391?

To solve for t, set v(t) equal to 8391: 3t + 12 = 8391; 3t = 8379; t = 2793

Now, the position function is equal to ∫v(t)dt = ∫ 3t + 12 dt = (3/2)t2 + 12t + C

Now, since the initial position is 44.5, C is 44.5; therefore, s(t) = (3/2)t2 + 12t + 44.5

To solve for the position, solve s(2793) = (3/2)27932 + 122793 + 44.5 = 1.5 7800849 + 33516 + 44.5 = 11701273.5 + 33516 + 44.5

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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

An eagle flies at a parabolic trajectory such that , where is in the height in meters and is the time in seconds. At what time will its velocity $v=-8 \frac{m}{s}$ ?

Answer

Take the derivative of the position function to obtain the velocity function.

$v(t)=\frac{\mathrm{d} }{\mathrm{d} t}h(t)=-4t$

We want to know the time when the velocity is -8. Substitute v into the equation to find t.

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Question

The position of a bike is described by the function . How long does it take the bike to reach a speed of ?

Answer

We need to find when . The velocity equation is the first derivative of the position equation. Taking the first derivative of the position equation gives

$v(t)=r'(t)=\frac{d}{dt}(2t^2)-\frac{d}{dt}(7t)+\frac{d}{dt}(6)$

Substituting gives

$t=\frac{19}{4}=4.75$

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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

A ball is thrown in the air, modeled by the function . At what time will the ball hit the ground?

Answer

To find the time when the ball hits the ground, set and solve for .

Separate each term and solve for t.

$t=\frac{3}{2}$

Since negative time does not exist, the only possible answer is $t=\frac{3}{2}$ .

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Question

Suppose the acceleration function of a biker going uphill at the start of a race is , where is in seconds. When will it take the biker to reach constant velocity?

Answer

Constant velocity means there is neither an increase or decrease in acceleration.

Substitute acceleration and solve for time.

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Question

Assume the acceleration due to gravity is

If you throw a ball straight up with an initial velocity of how many seconds will it take before the ball returns to your hand?

Answer

Write out the equation for the height of the ball

$h=-\frac{9.8}{2}t^{2}+9.8t$ .

You can arrive at this by starting with the information that the acceleration on the ball is the constant acceleration due to gravity and integrating twice. Making sure to solve for your constants along the way.

$v(t)=\int a(t)dt = \int -9.8 dt = -9.8t+v_{0}$

$x(t)=\int v(t)dt =\int -9.8t+v_{0}dt=-\frac{9.8}{2}t^{2}+v_{0}t+x_{0}$

Your initial velocity of and your initial position of will help write out this equation.

$x(t)=-\frac{9.8}{2}t^2+9.8t+0$

Then solve for the two values of for which the ball is at height . Those are seconds and seconds.

$\x(t)=-\frac{9.8}{2}t^2+9.8t+0=0\ \ -4.9t^{2}+9.8t=0\ \ t(-4.9t+9.8)=0\ \t=0,t=2$

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Question

A ball is thrown upwards at a speed of from a building. Assume gravity is .

Which of the following is closest to the time after the initial throw before the ball hits the ground?

Answer

If we approximate gravity as we can simplify into and use the quadratic formula to find the time at which the position of the ball is zero (the ball hits the ground).

$t=\frac{-2\pm \sqrt{(2)^2-4(-1)(20)}}{2(-1)}=1\pm\frac{\sqrt{4+80}}{-2}=1\mp\sqrt{21}$

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Question

Car A starts driving north from point O with an acceleration of . After 2 hours, Car B start driving north from point O with an acceleration of . How long will it take for Car B to catch up with Car A?

Answer

We know that car A's acceleration formula is $a_{A}(t)=10$ and we know that car B's acceleration formula is $a_{B}(t)=30$ . To solve this equation we must realize that the integral of acceleration is velocity and the integral of velocity is position. Therefore by taking the double integral of both acceleration functions, we can determine the point at which car B will catch up to car A.

Using the general integral formula,

$\int x^n=\frac{x^{n+1}}{n+1}+C$

we find that the velocity functions for both cars are $v_{A}(t)=10t$ and $v_{B}(t)=30t$ . Because the initial velocity of both cars is 0, .

Taking the integral of the velocity function using the generla integral formula once again, we find that the position functions of both cars is $S_{A}(t)=5t^2$ and $S_{B}(t)=15t^2$ . Since the initial position of both cars is equivalent, we can arbitrarily say that they start from a initial position 0, therefore making . We know that car A had a head start of 2 hours on car B. Now all we have to do is set both equations equal to each other and solve for .

$t=1+\sqrt{3} hours$

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Question

For this problem, the acceleration of gravity is simplified to $10 \frac{m}{s^2}$ .

I fire a cannon straight into the air. Assuming that the cannonball leaves the cannon at a velocity of $100 \frac{m}{s}$ , weighs kgs, and is fired from the ground (i.e. m), how long will it take before the cannonball reaches the ground again?

Answer

Initial velocity is given as 100 m/s and the acceleration due to gravity is $10\frac{m}{s^2}$ towards the Earth , or $-10 \frac{m}{s^2}$ .

We want to find how long it will take for the velocity of the cannonball to reach , so we set $v = 100 + a\cdot t = 0$ , where t is time and $a = -10 \frac{m}{s^2}$ .

So, , , .

Therefore, it takes ten seconds for the ball to reach a velocity of 0, and given that acceleration is uniform (i.e. a constant), we know that it will take the same amount of time to come down as it took to go up, or ten seconds. Therefore, the total time the cannonball spends in the air is,

seconds.

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Question

Given the instantaneous velocity and the position function, find the time at which the moving object reaches that instantaneous velocity.

Answer

We begin by finding the derivative of the position function using the power rule:

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

We then set the given instantaneous velocity equal to the velocity function and solve for t:

$24t+29=125\Rightarrow t=4$

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Question

At what time will a particle whose position can be described by have minimum acceleration?

Answer

We need to take the derivative of acceleration and set it equal to zero because we want to minimize acceleration. In total, this will be three derivatives

$\small x(t)\rightarrow v(t) \rightarrow a(t)\rightarrow \frac{\mathrm{d} }{\mathrm{d} t}a(t)$ .

Using the power rule on each term which states to multiply the coefficient by the exponent then decrease the exponent by one we get the following derivatives.

This will give us

, which gives us $\small t=.27$ and $\small t=-.27$ . Now let's use the second derivative test to see where our minimum is. $\small 480t$ is our second derivative, which is positive at $\small t=.27$ and negative at $\small t=-.27$ , so $\small t=.27$ is our minimum.

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Question

A perfectly spherical hot air balloon is being filled up. If the balloon is empty at the start and has a radius of 50 meters when fully inflated, how fast is the volume of the balloon increasing when its radius is 10 meters and increasing at a rate of $1\frac{m}{min}$ ?

Answer

In order to solve this problem, we must first know that the volume of a sphere is equivalent to $\frac{4}{3}\pi r^3$ .

In order to find the rate of which the volume of this spherical hot air balloon is increasing at, we must take the derivative of the volume equation with respect to time in order to find the change in volume with respect to time.

Using the power rule

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

we find that the dervative is

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

In the problem we are given the radius and rate of change of the radius, therefore by plugging those into the equation and solving for $\frac{dV}{dt}$ , we can find the rate at which the volume of the balloon is changing at.

Plugging and $\frac{dr}{dt}=1$ , we find that the volume of the baloon is increasing at a rate of $400 \pi \frac{m}{min}$ .

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Question

The position of a particle as a function of time is .

At what time is the particle at rest?

Answer

The particle is at rest when its velocity, i.e. the derivative of its position, is equal to 0.

Thus, we have to solve the equation

.

Using the Power Rule,

.

Thus, either or , leading to the solutions and .

Note: The Power Rule says that for a function

, $p'(x)=knx^{n-1}$ .

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