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Calculus 1 : Calculus

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Example Questions

Example Question #31 : Calculus

Given the following path:

$\left\{\begin{matrix}x(t)=t^3e^{t}\\ y(t)=t^4e^{t} \end{matrix}\right.$

What will be the velocity $\vec{v}$ of someone driving along that path?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To find the velocity we need to differentiate each of the components of the position with respect to time.

We note first that:

$(t^3e^{t})'=(t^3+3t^2)e^{t}$

$(t^4e^{t})'=(t^4+4t^3)e^{t}$ .

The velocity is then given by:

$\vec{v}=((t^3e^{t})',(t^4e^{t})')$ .

Using the above derivatives, we find the velocity:

$\vec{v}=((t^3+3t^2)e^{t},(t^4+4t^3)e^{t})$

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Example Question #32 : Calculus

A car is moving following the speed law:

${}\vect{v}=(t^2,t-1,t^3arctan(t))$

What is its acceleration?

Possible Answers:

$\left(2t,1,3t^2arctan(t)+\frac{t^3}{t^3+1}\right)$

$\left(2t,1,3t^2arctan(t^2)+\frac{t^3}{t^2+1}\right)$

No, they will never collide.

$\left(2t,1,3t^2arctan(t)+\frac{t^2}{t^2+1}\right)$

$\left(2t,1,3t^2arctan(t)+\frac{t^3}{t^2+1}\right)$

Correct answer:

$\left(2t,1,3t^2arctan(t)+\frac{t^3}{t^2+1}\right)$

Explanation:

To find the expression for the velocity, we need to differentiate each of the components with respect to time.

We have:

(t^2)'=2t

(t-1)'=1

$(t^3arctan(t))'=3t^2arctan(t)+\frac{t^3}{1+t^2}$

Collecting the components we obtain the expression for the velocity:

$\vec{v}=\left(2t,1,3t^2arctan(t)+\frac{t^3}{t^2+1}\right)$

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Example Question #33 : Calculus

The velocity of a particle is given by:

$\vec{V}=\left(\sqrt{3-t},\frac{e^{2t}}{t},ln(t+2)\right)$

When is the above expression defined?

Possible Answers:

(0,3]

(-2,3]

(-2,2)

(-2,3)

(-2,0)

Correct answer:

(0,3]

Explanation:

To have V defined for all t, we need to have each of the components defined as well.

First we see that for the first component to be defined , we need to have $3-t\ge 0$ .

For the second component we need to have $t\ne 0$ .

The last component must satisfy t+2 >0 .

This means that t must satisfy all these 3 restrictions.

This gives :

$\left\{\begin{matrix} t\le 3\\ t>-2 \\t>0 \end{matrix}\right.$

This means that $t\in (0,3]$

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Example Question #31 : Spatial Calculus

The position of a moving ant is given by:

$\vec{r}=(tsin(t),tcos(t),1)$

Find its speed.

Possible Answers:

$\sqrt{2+3t^2}$

t^2

$\sqrt{1+t^2}$

$\sqrt{1-2t^2}}$

$\sqrt{2+t^2}$

Correct answer:

$\sqrt{1+t^2}$

Explanation:

To find the speed, we need to find the velocity first. We are given the position vector. We need to differentiate to get the velocity vector and find its norm to find the speed.

To find the velocity we differentiate componentwise.

Using the Chain Rule we find the derivative to be:

$\vec{V}=(sin(t)+tcos(t),cos(t)-tsin(t),0)$

We take the norm of the above vector to get :

$||\vec{V}||=\sqrt{sin(t)+tcos(t))^2+(cos(t)-tsin(t))^2}=\sqrt{1+t^2}$

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Example Question #32 : Spatial Calculus

A car is moving through the following path:

$\vec{r}=(sin(t),sin(t),sin(t),sin(t))$

Find its speed at any given time.

Possible Answers:

2cos(t)

4sin(t)

4cos(t)

2sin(t)

2|cos(t))|

Correct answer:

2|cos(t))|

Explanation:

To find the speed, we need to find the velocity first. We need to differentiate the expression of the position vector componentwise with respect to time.

Knowing that (sin(t))=cos(t) .

We have the velocity given by:

$\vec{V}=(cos(t),cos(t),cos(t),cos(t))$ .

To find the speed we will have to find the norm of $\vec{V}$ .

$||\vec{V}||=\sqrt{cos^2(t)+cos^2(t)+cos^2(t)+cos^2(t)}$

and finally this gives:

$||\vec{V}||=\sqrt{4cos^2(t)}=2|cos(t)|$

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Example Question #36 : Calculus

Given the vector position:

$\vec{r}=(2+te^{-t},arctan(t)+1,arctan(2t))$

Find the expression of the velocity.

Possible Answers:

$\left((1-2t)e^{-t},\frac{1}{1+t^2},\frac{2}{4t^2+1}\right)$

$\left((1-t)e^{-t},\frac{1}{1+2t^2},\frac{2}{4t^2+1}\right)$

$\left((1-t)e^{-t},\frac{1}{1+t^2},\frac{2}{4t^2+1}\right)$

$\left((1-t)e^{-t},\frac{1}{1+t^2},\frac{1}{4t^2+1}\right)$

$\left((1-t)e^{-t},\frac{1}{1+t^2},\frac{1}{2t^2+1}\right)$

Correct answer:

$\left((1-t)e^{-t},\frac{1}{1+t^2},\frac{2}{4t^2+1}\right)$

Explanation:

To find the expression of the velocity vector, we must differentiate each of the position vector components with respect to time.

We have then:

$(te^{-t})'=(1-t)e^{-t}$

$(artan(t))'=\frac{1}{1+t^2}$

$(artan(2t))'=\frac{2}{4t^2+1}$

Finally collecting all these derivatives componentwise, we obtain the velocity.

$\left((1-t)e^{-t},\frac{1}{1+t^2},\frac{2}{4t^2+1}\right)$

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Example Question #33 : Spatial Calculus

Given the position, s(t)=5t^4 , find the velocity at time t=3 .

Possible Answers:

540

6075

405

243

Correct answer:

540

Explanation:

To find the velocity given the position function, take the derivative of the position function using the Power Rule.

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

Applying the Power Rule to our function we get the following.

s(t)=5t^4

$v(t)=\frac{ds}{dt}=4\cdot 5 t^{4-1}=20t^3$

Substitute t=3 to determine the velocity at this time.

v(t=3)=20(3)^3=540

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Example Question #33 : Velocity

Suppose the position function of an object is s(t)=ln(t) . What is the velocity of the object when $t=\frac{1}{3}$ ?

Possible Answers:

ln(3)

$ln(\frac{1}{3})$

$\frac{1}{3}$

Correct answer:

Explanation:

To find the velocity of s(t) , take the derivative of this function to find s'(t) .

s(t)=ln(t)

Recall that natural log has a special derivative.

$s'(t)=\frac{1}{t}$

Substitute $t=\frac{1}{3}$ and find the velocity at this time.

$s'\left(t=\frac{1}{3}\right)=\frac{1}{\frac{1}{3}}=3$

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Example Question #31 : Calculus

The position of an object is described by the function s(t)=3t+2 . What is the average velocity of this object from t=1 to t=6 ?

Possible Answers:

$\frac{17}{5}$

Correct answer:

Explanation:

Write the formula for average velocity.

$v_{avg}= \frac{s_{f}-s_{i}}{t_{f}-t_{i}}$

The variables $s_{f}$ is the final position, $s_{i}$ is the initial position, $t_{f}$ is the final time, and $t_{i}$ is the initial time.

Plug in $t_{i}=1$ and $t_{f}=6$ into s(t)=3t+2 to solve for initial and final position.

$s_{f}= 3(6)+2 =20$

$s_{i}=3(1)+2=5$

Substitute the known variables into the average velocity formula.

$v_{avg}= \frac{20-5}{6-1}=\frac{15}{5}=3$

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Example Question #32 : Spatial Calculus

Find the velocity function of a particle if the acceleration function is a(t)= -cos(x)+sin(x) .

Possible Answers:

-sin(x)+cos(x)

-sin(x)-cos(x)

sin(x)-cos(x)

sin(x)+cos(x)

Correct answer:

-sin(x)-cos(x)

Explanation:

To obtain the velocity function, integrate the acceleration function a(t)= -cos(x)+sin(x) .

$v(t)=\int(-cos(x)+sin(x))dx=-sin(x)-cos(x)$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #31 : Calculus

Example Question #32 : Calculus

Example Question #33 : Calculus

Example Question #31 : Spatial Calculus

Example Question #32 : Spatial Calculus

Example Question #36 : Calculus

Example Question #33 : Spatial Calculus

Example Question #33 : Velocity

Example Question #31 : Calculus

Example Question #32 : Spatial Calculus

All Calculus 1 Resources

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