AP Calculus AB - Spatial Calculus

The velocity of a particle is given by the function . What is the acceleration of the particle at time ?

Explanation

Acceleration of a particle can be found by taking the derivative of the velocity function with respect to time. Recall that a derivative gives the rate of change of some parameter, relative to the change of some other variable. When we take the derivative of velocity with respect to time, we are evaluating how velocity changes over time; i.e acceleration! This is just like finding velocity by taking the derivative of the position function.

$a(t)=\frac{d}{dt}v(t)$

$v(t)=\frac{d}{dt}p(t)$

Taking the derivative of the function

The acceleration function is

At time

The velocity of an object is given by the following equation:

$v(t)=\frac{3}{2}t^2-\frac{1}{3}t+16$

Find the equation for the acceleration of the object.

$a(t)=3t-\frac{1}{3}$

$a(t)=\frac{1}{3}t-3$

$a(t)=\frac{1}{2}t-16$

$a(t)=16t+\frac{1}{3}$

Explanation

Acceleration is the derivative of velocity, so in order to find the equation for the object's acceleration, we must take the derivative of the equation for its velocity:

$v(t)=\frac{3}{2}t^2-\frac{1}{3}t+16$

We will use the power rule to find the derivative which states:

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

$a(t)=v'(t)=\frac{3}{2}\cdot 2t^1-\frac{1}{3}t^0=3t-\frac{1}{3}$

Suppose a particle travels in a circular motion in the xy-plane with

$x(t) = R \cos(\omega t)$

$y(t) = R \sin(\omega t)$

for some constants $\omega, R$ . Notice that this is circular because .

What is the total magnitude of the acceleration, $a = \sqrt{(x'')^2 + (y'')^2}$ ?

$R \omega^2$

$2R\omega^2$

$R^2 \omega^4$

$2 \pi R \omega$

Explanation

We know that and so the acceleration components are:

$x''(t) = - R \omega^2 \cos(\omega t)$

$y''(t) = - R \omega^2 \sin(\omega t)$

Plugging these into the formula for the acceleration and again recognizing that $\sin^2(\theta) + \cos^2(\theta) = 1$ , we get $a^2 = R^2 \omega^4$ , or, after the square root, $a = R \omega^2$ .

The velocity of a particle is given by the function $v(t)=4e^{cos(3t)}$ . What is its acceleration function?

$-12sin(3t)e^{cos(3t)}$

$-12e^{cos(3t)}$

$-12cos(3t)e^{cos(3t)}$

$12sin(3t)e^{cos(3t)}$

$12cos(3t)e^{cos(3t)}$

Explanation

Acceleration is given by the time derivative of velocity.

In this case we will use the rules that the derivative of

is $e^u \cdot \frac{du}{dx}$

and the derivative of

is $-sin(u)\cdot \frac{du}{dx}$ .

Applying this knowledge we can find the acceleration function to be the following.

$a(t)=v'(t)=\frac{d}{dt}(4e^{cos(3t)})$

$a(t)=-12sin(3t)e^{cos(3t)}$

Given that the position function of a paper airplane is and its initial veloctiy is $6\frac{ft}{sec}$ find its acceleration at seconds.

$-4 \frac{ft}{s^2}$

$4 \frac{ft}{s^2}$

$-5 \frac{ft}{s^2}$

$5 \frac{ft}{s^2}$

None of the above.

Explanation

In order to solve this proble, it must first be realized that the derivative of a position function is the velocity function and the derivative of the velocity function becomes the acceleration function. In this problem we are given the position function and asked to find the acceleration after a certain amount of time. By taking the double derivative of the function and plugging in the time, we will be able to find the acceleration of the paper plane at that time. In order to take the derivative of the position function, we must first know the power rule, $\frac{d}{dx}x^n=nx^{n-1}$ .

Applying the power rule to the position function, we find the first derivative to be .

Applying the power rule a second time to the veloctiy function, we find the second derivative to be .

Now, by simply plugging seconds, we find that the acceleration of the paper airplane at that time is $-4\frac{ft}{s^2}$ .

Find the position at given the following velocity function.

Explanation

To solve, simply integrate to find , the position function, and then plug in .

$\int_{0}^{2}v(t) \ dt=\int_{0}^{2}2t+3 \ dt=t^2+3t \Big |_{0}^{2}=(2^2+3(2))-(0^2+3(0))=4+6=10$

Given that the velocity of a bird flying is given as , calculate how far the bird has traveled from to . The original position of the bird is unknown.

None of the above.

Explanation

To solve this problem, it is important to understand that the derivative of position is velocity. Thus by integrating the velocity function, we will be able to obtain the posiiton function. Because the problem asks us to find the distance the bird travels from to , we integrate from 0 to 3. The general formula for integration is

$\int x^{n}=\frac{x^{n+1}}{n+1}+C$ where C can be any real number.

Integrating the velocity function gives us

, where C is any real number.

However, the problem asks us the position change of the bird from to , therefore the position of the bird is at and at . By subtracting, we get that the distance travelled by the bird during this time interval is .

The velocity of a particle is given by the function $v(t)=3t^2\cos(t^3)-8t\sin(t^2)$ . How far does the particle travel over the interval of time ?

Explanation

Velocity is the time derivative of position, and by that token position can be found by integrating a known velocity function with respect to time:

$p(t)=\int v(t)dt$

Now, if this integral were to be taken over an interval of time , this will give a finite value, a change in position, i.e. a distance travelled:

$d=\int_a^b v(t)dt$

For the velocity function

$v(t)=3t^2\cos(t^3)-8t\sin(t^2)$

The distance travelled can be found via knowledge of the following derivative properties:

Trigonometric derivative:

$d[\sin(u)]=\cos(u)du$

$d[\cos(u)]=-\sin(u)du$

The distance travelled over is:

$d=\sin(t^3)+4\cos(t^2)\Big|_2^5$

$d=(\sin(5^3)+4\cos(5^2))-(\sin(2^3)+4\cos(2^2))$

The acceleration of a particle is given by the function $a(t)=5(1-e^{-2t})$ . If the particle has an initial velocity of zero, how far does it travel on the interval ?

Explanation

Begin by finding the velocity function, which is the integral of the acceleration function with respect to time.

For acceleration

$a(t)=5(1-e^{-2t})$

The integral of a constant is simpy , and the integral of an exponential function $e^{bt}$ is $\frac{1}{b}e^{bt}$ .

The velocity function can be found using these properties to be

$v(t)=5(t+\frac{1}{2}e^{-2t})+C$

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

$v(0)=5(0+\frac{1}{2}e^{-2(0)})+C=0$

$\frac{1}{2}+C=0$

$C=-\frac{1}{2}$

This gives the complete function

$v(t)=5(t+\frac{1}{2}e^{-2t})-\frac{1}{2}$

Now to find distance:

Distance traveled over an interval of time can be found by integrating the velocity function over this interval:

$d=\int_a^b v(t)dt$

Over the inteval

$d=\int_0^4 (5(t+\frac{1}{2}e^{-2t})-\frac{1}{2})dt$

Note that the integral of a function is $\frac{at^{n+1}}{n+1}$ .

$d= (\frac{5t^2}{2}-\frac{t}{2}-\frac{5e^{-2t}}{4})|_0^4$

$d=(\frac{80}{2}-\frac{4}{2}-\frac{5e^{-8}}{4})-(0-0-\frac{5}{4})$

The velocity of a car in miles per hour can be described with the function

$v(t)=7t^{\frac{1}{3}}+12t$ .

How many miles did the car travel between and hours?

$456.75\ miles$

$479.25\ miles$

$468\ miles$

$468.75\ miles$

There is not enough information to determine the solution.

Explanation

We know that distance is the difference of positions, so we need a position function. We can find this by integrating the velocity function.

Recall to integrate this particular function we use,

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

$p(t)=\int v(t)dt=\int (7t^{\frac{1}{3}}+12t)dt$

$\p(t)=\frac{7}{\frac{1}{3}+1}t^{\frac{1}{3}+1}+\frac{12}{2}t^2+c\ \ \p(t)=\frac{7}{\frac{4}{3}}t^\frac{4}{3}+6t^2+c\ \ \p(t)=\frac{21}{4}t^\frac{4}{3}+6t^2+c$