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Calculus 1 : Spatial Calculus

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Example Questions

Example Question #141 : Spatial Calculus

A particle's position is defined by $s(t) = 4e^{t^{2}\sin(t)} + \2t\ln(t)$ $\displaystyle s(t) = 4e^{t^{2}\sin(t)} + \2t\ln(t)$

What is the instantaneous velocity of the particle at $t = \pi$ $\displaystyle t = \pi$?

Possible Answers:

$\displaystyle 0$

$2 - 4\pi^{2} + \ln(\pi)$ $\displaystyle 2 - 4\pi^{2} + \ln(\pi)$

$2 + 4\pi + 2\ln(\pi)$ $\displaystyle 2 + 4\pi + 2\ln(\pi)$

$2\pi^2 - \ln(\pi)$ $\displaystyle 2\pi^2 - \ln(\pi)$

$6 + 2\ln(\pi)$ $\displaystyle 6 + 2\ln(\pi)$

Correct answer:

$2 - 4\pi^{2} + \ln(\pi)$ $\displaystyle 2 - 4\pi^{2} + \ln(\pi)$

Explanation:

To find the velocity function, take the derivative of the position funtion.

$s'(t) = v(t) = 4e^{t^2\sin(t)}(t^2\cos(t) + 2t\sin(t)) + 2 + 2\ln(t)$ $\displaystyle s'(t) = v(t) = 4e^{t^2\sin(t)}(t^2\cos(t) + 2t\sin(t)) + 2 + 2\ln(t)$

Now plug in $\pi$ $\displaystyle \pi$ to find the velocity:

$v(\pi) = 4e^0(-\pi^2 + 0) + 2 + 2\ln(\pi) = 2 - 4\pi^2 + 2\ln(\pi)$ $\displaystyle v(\pi) = 4e^0(-\pi^2 + 0) + 2 + 2\ln(\pi) = 2 - 4\pi^2 + 2\ln(\pi)$

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Example Question #142 : Spatial Calculus

If $\small \small s(t)=e^{t^2}$ $\displaystyle \small \small s(t)=e^{t^2}$ is the position of a particle, what is the velocity at time $\small t=4$ $\displaystyle \small t=4$?

Possible Answers:

$\small \small \small \small v(4)=2e^{16}$ $\displaystyle \small \small \small \small v(4)=2e^{16}$

$\small \small v(4)=8e^{16}$ $\displaystyle \small \small v(4)=8e^{16}$

$\small \small \small v(4)=16e^{16}$ $\displaystyle \small \small \small v(4)=16e^{16}$

$\small \small \small v(4)=8e^{8}$ $\displaystyle \small \small \small v(4)=8e^{8}$

Correct answer:

$\small \small v(4)=8e^{16}$ $\displaystyle \small \small v(4)=8e^{16}$

Explanation:

If we have the position $\small s(t)$ $\displaystyle \small s(t)$, then the velocity is the derivative of the position:

$\small v(t)=s'(t)=2te^{t^2}$ $\displaystyle \small v(t)=s'(t)=2te^{t^2}$

by using the chain rule. So we plug in $\small t=4$ $\displaystyle \small t=4$ to get

$\small \small v(4)=8e^{16}$ $\displaystyle \small \small v(4)=8e^{16}$

as the velocity.

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Example Question #141 : Calculus

If the acceleration of a particle at time $\small t$ $\displaystyle \small t$ is $\small a(t)=2t+1$ $\displaystyle \small a(t)=2t+1$, and the initial velocity (i.e., at $\small t=0$ $\displaystyle \small t=0$) is $\small v(0)=2$ $\displaystyle \small v(0)=2$, what is the velocity at $\small t=2$ $\displaystyle \small t=2$?

Possible Answers:

Not enough information to find the velocity at $\small t=2$ $\displaystyle \small t=2$.

$\small v(2)=2$ $\displaystyle \small v(2)=2$

$\small v(2)=8$ $\displaystyle \small v(2)=8$

$\small v(2)=6$ $\displaystyle \small v(2)=6$

Correct answer:

$\small v(2)=8$ $\displaystyle \small v(2)=8$

Explanation:

This is simply an application of the fundamental theorem of calculus, since we know that $\small a(t)=v'(t)$ $\displaystyle \small a(t)=v'(t)$:

$\small \int_0^2 a(t)dt = \int_0^2 v'(t)dt=v(2)-v(0)$ $\displaystyle \small \int_0^2 a(t)dt = \int_0^2 v'(t)dt=v(2)-v(0)$

So we have

$\small v(2)=v(0)+\int_0^2 (2t+1)dt$ $\displaystyle \small v(2)=v(0)+\int_0^2 (2t+1)dt$

So let's solve for $\small \int_0^2 (2t+1)dt$ $\displaystyle \small \int_0^2 (2t+1)dt$:

$\small \small \int_0^2 (2t+1)dt=t^2+t|_0^2=2^2+2-(0^2-0)=6$ $\displaystyle \small \small \int_0^2 (2t+1)dt=t^2+t|_0^2=2^2+2-(0^2-0)=6$

So the velocity at $\small t=0$ $\displaystyle \small t=0$ is

$\small \small v(2)=v(0)+\int_0^2 (2t+1)dt=2+6=8$ $\displaystyle \small \small v(2)=v(0)+\int_0^2 (2t+1)dt=2+6=8$

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Example Question #143 : Spatial Calculus

If the velocity is $\small \small v(t)=e^{\sin t}$ $\displaystyle \small \small v(t)=e^{\sin t}$, then what is the acceleration function, $\small a(t)$ $\displaystyle \small a(t)$?

Possible Answers:

$\small a(t)=\cos te^{\sin t}$ $\displaystyle \small a(t)=\cos te^{\sin t}$

$\small \small a(t)=-\cos te^{\sin t}$ $\displaystyle \small \small a(t)=-\cos te^{\sin t}$

$\small \small a(t)=-\sin te^{\cos t}$ $\displaystyle \small \small a(t)=-\sin te^{\cos t}$

$\small \small a(t)=\sin te^{\cos t}$ $\displaystyle \small \small a(t)=\sin te^{\cos t}$

Correct answer:

$\small a(t)=\cos te^{\sin t}$ $\displaystyle \small a(t)=\cos te^{\sin t}$

Explanation:

To get the acceleration from $\small \small v(t)=e^{\sin t}$ $\displaystyle \small \small v(t)=e^{\sin t}$, we take the derivative to get:

$\small a(t)=v'(t)=\cos te^{\sin t}$ $\displaystyle \small a(t)=v'(t)=\cos te^{\sin t}$

with the use of the chain rule

$f(g(x))'=f'(g(x))\cdot g'(x)$ $\displaystyle f(g(x))'=f'(g(x))\cdot g'(x)$.

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Example Question #144 : Spatial Calculus

A car's position is represented by the function

$\displaystyle x(t)=2t^2+4t+1$.

From this information, find the car's velocity when t=6 $\displaystyle t=6$.

Possible Answers:

$\displaystyle 16$

$\displaystyle 4$

$\displaystyle 8$

$\displaystyle 28$

$\displaystyle 97$

Correct answer:

$\displaystyle 28$

Explanation:

The given equation is used to find position. Based on what we know about derivatives, if we take the derivative of position with respect to time we will be finding the change in position over time, which is velocity. So the first step is to take the derivative. We will get

$\displaystyle x'(t)=v(t)=4t+4$

to represent our velocity. Next, you plug in the given t value of t=6 to find the velocity at this step in time.

$\displaystyle v(6)=4(6)+4=28$

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Example Question #145 : Spatial Calculus

An object's position is described by the given function:

$\displaystyle x(t)=16t+4$.

Given this information is the object's velocity constant or changing? What is the velocity when t=2 $\displaystyle t=2$?

Possible Answers:

$\displaystyle Constant; 0$

$\displaystyle Varying; 16$

$\displaystyle Constant; 16$

$\displaystyle Varying; 36$

$\displaystyle Constant; 36$

Correct answer:

$\displaystyle Constant; 16$

Explanation:

$\displaystyle x'(t)=v(t)=16$

to represent our velocity. Since our function for velocity is a constant, we know that the velocity will always be 16. To confirm this, we can take the derivative of our velocity to find acceleration. The derivative of a constant is 0, therefore, there is no acceleration or change in velocity, thus velocity is constant at 16.

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Example Question #146 : Spatial Calculus

A bug's position is represented by the function $\displaystyle x(t)=t^3-2t^2+7$. Using the given information, find the velocity when t=5 $\displaystyle t=5$.

Possible Answers:

$\displaystyle 55$

$\displaystyle 0$

$\displaystyle 65$

$\displaystyle -55$

$\displaystyle 82$

Correct answer:

$\displaystyle 55$

Explanation:

$\displaystyle x'(t)=v(t)=3t^2-4t$

to represent our velocity. Next, you plug in the given t value of t=5 to find the velocity at this step in time.

$\displaystyle v(5)=3(5)^2-4(5)=75-20=55$

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Example Question #141 : Velocity

If f(x) $\displaystyle f(x)$ models the position of a frisbee as a function of time. Find the equation which models the velocity of the frisbee.

$\displaystyle f(x)=x^3+5x^2-4x$

Possible Answers:

$\displaystyle f'(x)=3x^2-4x+10$
$\displaystyle f'(x)=3x^2+10x$

$\displaystyle f'(x)=6x^2+5x$
$\displaystyle f'(x)=x^2+5x-4$

$\displaystyle f'(x)=3x^2+10x$
$\displaystyle f'(x)=6x+10$

$\displaystyle f'(x)=3x^2+10x-4$$\displaystyle f'(x)=3x^2+10x-4$$\displaystyle f'(x)=3x^2+10x-4$

Correct answer:

$\displaystyle f'(x)=3x^2+10x-4$$\displaystyle f'(x)=3x^2+10x-4$$\displaystyle f'(x)=3x^2+10x-4$

Explanation:

Recall that velocity if the first derivative of position, so to find the equation which models the velocity, we need to take the derivative.

Recall that for any polynomial we can take the derivative as follows:

$\displaystyle f(x)=x^n$

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

$\displaystyle f(x)=x^n$

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

So our original function:

$\displaystyle f(x)=x^3+5x^2-4x$

Becomes:

$\displaystyle f'(x)=3x^2+10x-4$

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Example Question #147 : Spatial Calculus

If f(x) $\displaystyle f(x)$ models the position of a frisbee as a function of time, find the velocity of the frisbee after $\displaystyle 12$ seconds.

$\displaystyle f(x)=x^3+5x^2-4x$

Possible Answers:

$548\frac{m}{s}$ $\displaystyle 548\frac{m}{s}$

$144\frac{m}{s}$ $\displaystyle 144\frac{m}{s}$

$484\frac{m}{s}$ $\displaystyle 484\frac{m}{s}$

$44.8\frac{m}{s}$ $\displaystyle 44.8\frac{m}{s}$

Correct answer:

$548\frac{m}{s}$ $\displaystyle 548\frac{m}{s}$

Explanation:

Recall that velocity if the first derivative of position, so to find the equation which models the velocity, we need to take the derivative.

Recall that for any polynomial we can take the derivative as follows:

$\displaystyle f(x)=x^n$

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

So our original function:

$\displaystyle f(x)=x^3+5x^2-4x$

Becomes:

$\displaystyle f'(x)=3x^2+10x-4$

Then, to find the velocity after 12 seconds, find f'(12)

$f'(12)=3(12)^2+10(12)-4=(3\cdot 144)+120-4=548\frac{m}{s}$ $\displaystyle f'(12)=3(12)^2+10(12)-4=(3\cdot 144)+120-4=548\frac{m}{s}$

So our answer is 448 meters per second. Perhaps a bit fast for a frisbee, but it is the correct answer in this case!

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Example Question #148 : Spatial Calculus

If p(t) $\displaystyle p(t)$ models the position of an electron as a function of time. Find v(t) $\displaystyle v(t)$, the velocity function of the electron:

$\displaystyle p(t)=4cos(t)+4$

Possible Answers:

$\displaystyle v(t)=4sin(t)$

$\displaystyle v(t)=-4sin(t)$

$\displaystyle v(t)=-sin(4t)$

$\displaystyle v(t)=-4sin(t)+4t$

Correct answer:

$\displaystyle v(t)=-4sin(t)$

Explanation:

If p(t) models the position of an electron as a function of time. Find v(t), the velocity function of the electron:

$\displaystyle p(t)=4cos(t)+4$

Recall that velocity is the derivative of position. To find v(t), we need to find p'(t).

Remember that the derivative of cosine is negative sine, so we get...

$\displaystyle p(t)=4cos(t)+4$

$\displaystyle p'(t)=-4sin(t)$

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Calculus 1 : Spatial Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #141 : Spatial Calculus

Example Question #142 : Spatial Calculus

Example Question #141 : Calculus

Example Question #143 : Spatial Calculus

Example Question #144 : Spatial Calculus

Example Question #145 : Spatial Calculus

Example Question #146 : Spatial Calculus

Example Question #141 : Velocity

Example Question #147 : Spatial Calculus

Example Question #148 : Spatial Calculus

All Calculus 1 Resources

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