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Calculus 1 : Calculus

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Example Questions

Example Question #171 : Spatial Calculus

A molecule moves with a very odd path, following the function y=sin(cos(tan(x))) .

Find the velocity of the molecule at x=0 .

Possible Answers:

Undefined

cos(cos(1))

Correct answer:

Explanation:

This problem requires using the chain rule which states

$f(g(x))'=f'(g(x))\cdot g'(x)$ several times.

First we take the derivative of the outside, and move inward.

$\frac{\mathrm{d} }{\mathrm{d} x}sin(cos(tan(x)))=$

cos(cos(tan(x))*-sin(tan(x))*sec^2(x)

At x=0 , we get

cos(cos(tan(0))*-sin(tan(0))*sec^2(0)= 0

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Example Question #171 : Calculus

The position of a particle is given by the function x(t)=sin(3t^2+2t+6)

What is the particle's velocity at time t=3 ?

Possible Answers:

39cos(39)

39cos(20)

20cos(39)

20cos(20)

cos(39)

Correct answer:

20cos(39)

Explanation:

Velocity can be found by taking the time derivative of the position function:

$v(t)=\frac{d}{dt}x(t)=(6t+2)cos(3t^2+2t+6)$

Therefore, at t=3

v(3)=(18+2)cos(27+6+6)

v(3)=20cos(39)

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Example Question #171 : How To Find Velocity

Find the acceleration of a particle whose position function is given by the formula:

$x(t)=t^2+cos(2t)+e^{3t}$

Possible Answers:

$2-4t^2cos(2t)+9t^2e^{3t}$

$2+4cos(2t)+9e^{3t}$

$2t-2sin(2t)+3e^{3t}$

$2-4sin(2t)+9e^{3t}$

$2-4cos(2t)+9e^{3t}$

Correct answer:

$2-4cos(2t)+9e^{3t}$

Explanation:

Acceleration can be found as the second time derivative of position:

$a(t)=\frac{d^2}{dt^2}x(t)$

Therefore, as position is

$x(t)=t^2+cos(2t)+e^{3t}$

Acceleration is:

$a(t)=2-4cos(2t)+9e^{3t}$

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Example Question #172 : Velocity

Find the velocity at t=2 for a particle whose position is given by the function:

$x(t)=t^3+tsin(\frac{\pi}{2}t)$

Possible Answers:

$12-2\pi$

$12+\pi$

$12-\pi$

Correct answer:

$12-\pi$

Explanation:

Velocity is the time derivative of position:

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

For a position function:

$x(t)=t^3+tsin(\frac{\pi}{2}t)$

The velocity function is:

$v(t)=3t^2}+\frac{\pi}{2}tcos(\frac{\pi}{2}t)+sin(\frac{\pi}{2}t)$

Therefore

$v(2)=3(4)+\frac{\pi}{2}(2)cos(\frac{\pi}{2}(2))+sin(\frac{\pi}{2}(2))$

$v(2)=12-\pi$

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Example Question #175 : Spatial Calculus

A particle's position is given by the function $x(t)=\frac{sin(3t)}{2+cos(t)}$

What is its velocity at time $t=\pi$ ?

Possible Answers:

$-2\pi$

$2\pi$

Correct answer:

Explanation:

Velocity can be found as the time derivative of the position function:

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

If the postion function is

$x(t)=\frac{sin(3t)}{2+cos(t)}$

Then the velocity function is subsequently:

$v(t)=\frac{3cos(3t)(2+cos(t))+sin(3t)sin(t)}{(2+cos(t))^2}$

Therefore at time $t=\pi$

$v(\pi)=\frac{3(-1)(2-1)+0(0)}{1^2}$

$v(\pi)=-3$

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Example Question #176 : Spatial Calculus

The position of a particle is given by the function $x(t)=t^2e^{2t}sin(3t)$ .

What is its velocity at time t=2 ?

Possible Answers:

12e^4cos(6)+4e^4sin(6)

12e^4cos(6)+8e^4sin(6)

12e^4cos(6)-12e^4sin(6)

12e^4cos(6)+12e^4sin(6)

12e^4cos(6)+6e^4sin(6)

Correct answer:

12e^4cos(6)+12e^4sin(6)

Explanation:

Velocity is the time derivative of position:

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

If position is:

$x(t)=t^2e^{2t}sin(3t)$

Then the velocity is:

$v(t)=2te^{2t}sin(3t)+t^2(2e^{2t}sin(3t)+3e^{2t}cos(3t))$

$v(t)=3t^2e^{2t}cos(3t)+2t^2e^{2t}sin(3t)+2te^{2t}sin(3t)$

$v(2)=12e^{4}cos(6)+8e^{4}sin(6)+4e^{4}sin(6)$

v(2)=12e^4cos(6)+12e^4sin(6)

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Example Question #177 : Spatial Calculus

Find the velocity function of a car if the position function of the car is given by

$s(t)=4t^4+\cos(t)+2t$

Possible Answers:

$8t^3-\sin(t)+2$

$16t^3-\sin(t)+2$

$16t^3+\sin(t)+2$

$16t^3-\sin(t)+2t$

Correct answer:

$16t^3-\sin(t)+2$

Explanation:

The first derivative of the position function - the velocity function - is

$s'(t)=v(t)=16t^3-\sin(t)+2$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #178 : Spatial Calculus

A basketball has the following position equation:

s(t)=-3t^2+20

What is the velocity of the basketball at time t=1 ?

Possible Answers:

Correct answer:

Explanation:

To find the velocity equation of the basketball, we need to take the derivative of the position equation:

s'(t)=v(t)=-6t

We used the following rule to find the derivative:

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

Now, since we were asked to find the velocity at t=1, plug in t=1 into the velocity equation:

v(1)=-6

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Example Question #174 : Calculus

The position of a particle is given by the function $x(t)=3t+t^2e^{-2t}$ .

What is its velocity at time t=3 ?

Possible Answers:

$3+3e^{-6}$

$-3e^{-6}$

$3-9e^{-6}$

$3-12e^{-6}$

$3e^{-6}$

Correct answer:

$3-12e^{-6}$

Explanation:

Velocity is the time derivative of position.

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

For the position function:

$x(t)=3t+t^2e^{-2t}$

For this function we will need to use the rules of exponents, the product rule, and the power rule to differentiate.

Rules of exponents: $e^u\rightarrow e^u\cdot \frac{du}{dx}$

Product Rule: $[f(x)g(x)]'\rightarrow f(x)g'(x)+g(x)f'(x)$

Power Rule: $x^n \rightarrow nx^{n-1}$

Applying these rules we find the velocity function to be:

$v(t)=3+2te^{-2t}-2t^2e^{-2t}$

$v(3)=3+2(3)e^{-2(3)}-(2)3^2e^{-2(3)}$

$v(3)=3+6e^{-6}-18e^{-6}$

$v(3)=3-12e^{-6}$

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Example Question #180 : Spatial Calculus

Find the velocity function of a train if its position function is $s(t)=7t^{2}-19$ .

Possible Answers:

$v(t)=7t^{2}$

v(t)=14

v(t)=14t

v(t)=-19

$v(t)=7t^{2}-19$

Correct answer:

v(t)=14t

Explanation:

The velocity function is the first derivative of the position function, or

$v(t)=s'(t)=\frac{d}{dt}(s(t))$ .

We will use the power rule, $\frac{d}{dt}(t^{c})=ct^{c-1}$ , where is a constant,

and the constant rule, $\frac{d}{dt}(c)=0$ , where is a constant, to find the derivative of this position function.

For this problem:

$s(t)=7t^{2}-19$

$v(t)=\frac{d}{dt}(s(t))=7\frac{d}{dt}(t^{2})-\frac{d}{dt}(19)$

v(t)=7*2t

v(t)=14t

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #171 : Spatial Calculus

Example Question #171 : Calculus

Example Question #171 : How To Find Velocity

Example Question #172 : Velocity

Example Question #175 : Spatial Calculus

Example Question #176 : Spatial Calculus

Example Question #177 : Spatial Calculus

Example Question #178 : Spatial Calculus

Example Question #174 : Calculus

Example Question #180 : Spatial Calculus

All Calculus 1 Resources

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