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Calculus 1 : Spatial Calculus

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Example Question #3 : Derivatives

What is the velocity function when the position function is given by

p(t)=t^2+t .

Possible Answers:

v(t)=2

v(t)=2t+1

v(t)=t^2

v(t)=t^2+1

v(t)=t^3

Correct answer:

v(t)=2t+1

Explanation:

To find the velocity function, we need to find the derivative of the position function.

So lets take the derivative of p(t)=t^2+t with respect to .

The derivative of t^2 is because of Power Rule:

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

$p(t)=t^2 \rightarrow p'(t)=2t^{2-1}=2t$

The derivative of is due to Power Rule

$p(t)=t \rightarrow p'(t)=1t^{1-1}=1t^0=1(1)=1$

So...

v(t)=2t+1

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Example Question #21 : How To Find Velocity

What is the velocity when the velocity function is v(t)=4t+3 at time t=2 ?

Possible Answers:

v(2)=11

v(2)=3

v(2)=7

v(2)=4

v(2)=2

Correct answer:

v(2)=11

Explanation:

To find the velocity at t=2 we just plug into the veloctiy function

$v(2)=4\cdot 2+3$

v(2)=8+3

v(2)=11

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Example Question #22 : How To Find Velocity

Find the velocity function when the position function is

p(t)=3t^2+4 .

Possible Answers:

v(t)=3t

v(t)=4

v(t)=6t

v(t)=4t

v(t)=6

Correct answer:

v(t)=6t

Explanation:

To find the veloctiy function, we need to take the derivative of the position function

So p(t)=3t^2+4 would turn into v(t)=6t

because of the Power Rule as well as the rule that the derivative of a constant is .

$p(t)=a^n \rightarrow v(t)=na^{n-1}$

$v(t)=2\cdot 3t^1+0=6t$

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Example Question #23 : How To Find Velocity

What is the velocity when the position function is

p(t)=t^3+4t .

Possible Answers:

v(t)=3t^2

v(t)=3t+4

v(t)=3t^3

v(t)=3t^2+4

v(t)=3t

Correct answer:

v(t)=3t^2+4

Explanation:

To find the velocity function, we need to take the derivative of the position function in relation to

p(t)=t^3+4t

changes into

v(t)=3t^2+4

using the Power Rule

$p(t)=a^n \rightarrow v(t)=na^{n-1}$

$v(t)=3 \cdot 1t^{3-1}+4t^{1-1}=3t^2+4t^0=3t^2+4$ .

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Example Question #24 : How To Find Velocity

What is the velocity when t=6 when the velocity function is

v(t)=t^2+3t ?

Possible Answers:

v(6)=6

v(6)=36

v(6)=48

v(6)=54

v(6)=18

Correct answer:

v(6)=54

Explanation:

To find the velocity at time t=6 we plug in for

v(t)=t^2+3t

becomes

$v(6)=6^2+3\cdot 6$

v(6)=36+18

v(6)=54 .

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Example Question #25 : How To Find Velocity

What is the velocity function when the position function is

p(t)=t^4+t^2+2t ?

Possible Answers:

v(t)=4t^3+2t^2+2

v(t)=4t^3

v(t)=4t^3+2t+2

v(t)=4t

v(t)=4t^2+2t

Correct answer:

v(t)=4t^3+2t+2

Explanation:

To find the velocity function we need to take the derivative of the position function

So p(t)=t^4+t^2+2t turns into v(t)=4t^3+2t+2 because of power rule.

$p(t)=a^n \rightarrow v(t)=na^{n-1}$

$v(t)=4t^{4-1}+2t^{2-1}+2t^{1-1}=4t^3+2t+2$

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Example Question #26 : How To Find Velocity

If h(t) models the distance of a projectile as a function of time, find the velocity of the projectile at t=15 .

$h(t)=\frac{4x^3}{3}+\frac{4.5x^2}{3}+3x-7$

Possible Answers:

$48\frac{m}{s}$

$64\frac{m}{s}$

$948\frac{m}{s}$

$65\frac{m}{s}$

$963\frac{m}{s}$

Correct answer:

$948\frac{m}{s}$

Explanation:

We are given a function dealing with distance and asked to find a velocity. recall that velocity is the first derivative of position. Find the first derivative of h(t) and evaluate at t=15.

$h(t)=\frac{4t^3}{3}+\frac{4.5t^2}{3}+3t-7$

h'(t)=4t^2+3t+3

h'(15)=4*15^2+3*15+3=900+45+3=948

$948\frac{m}{s}$

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Example Question #27 : How To Find Velocity

Consider the following position function:

p(t)=6t^4-5t^3+6t-18

Find the velocity of a paper airplane after seconds whose position is modeled by {}p(t) .

Possible Answers:

$343 \frac{m}{s}$

$519 \frac{m}{s}$

$51.9 \frac{m}{s}$

$19 \frac{m}{s}$

$5.19 \frac{m}{s}$

Correct answer:

$519 \frac{m}{s}$

Explanation:

Recall that velocity is the first derivative of position.

To find the velocity after 3 seconds, we need p'(3).

p(t)=6t^4-5t^3+6t-18

p'(t)=24t^3-15t^2+6

$p'(3)=24\cdot3^3-15\cdot3^2+6=519$

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Example Question #28 : How To Find Velocity

The velocity of a driver is given by the following vector:

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

What velocity will she or he reach in the long run?

Possible Answers:

$\left(1,0,-\frac{\pi}{2}\right)$

$\left(1,0,\frac{\pi}{2}\right)$

$\left(1,1,\frac{\pi}{2}\right)$

$(1,0,\pi)$

$\left(1,0,\frac{\pi}{4}\right)$

Correct answer:

$\left(1,0,\frac{\pi}{2}\right)$

Explanation:

To find the velocity for the long run, we need to find the limit as t gets bigger and bigger.

Note that to find this limit, we need to take the limit componentwise.

We know the following:

$\lim_{t\rightarrow \infty}\frac{t^2}{t^2+1}=1$

$\lim_{t\rightarrow \infty}e^{-t^2}=0$

$\lim_{t\rightarrow \infty}artan(t^2)=\frac{\pi}{2}$ .

Therefore as t becomes bigger we will reach the velocity given by:

$\left(1,0,\frac{\pi}{2}\right)$

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Example Question #29 : How To Find Velocity

Two drivers have the following position vectors:

$\vec{OM_{1}}=(cos(t),sin(t))$

$\vec{OM_{2}}=(cos(2t),sin(2t))$

Wich driver is driving faster?

Possible Answers:

The first driver.

They have the same velocity.

We can't tell from the given information.

The second driver.

They follow the same path and have the same velocity.

Correct answer:

The second driver.

Explanation:

To find who is driving faster, we need to compare the norm of their velocities:

We will call $\vec{V_{1}},\vec{V_{2}}$ the velocities of the first and the second drivers respectively.

Using the Chain Rule on the position function we have the following velocities,

$\vec{V_{1}}=(-sin(t),cos(t))$ .

$\vec{V_{2}}=(-2sin(2t),2cos(2))$ .

Now we find the speed of both drivers.

$v_{1}=\sqrt{sin^2(t)+cos^2(t)}=1$

Use the trigonometric identity to find velocity two.

$v_{2}=\sqrt{4sin^2(t)+4cos^2(t)}=\sqrt{4\left( \frac{1}{2}-\frac{1}{2}cos(2x)\right)+4\left(\frac{1}{2}+\frac{1}{2}cos(2x)\right)}$

$v_2=\sqrt{2-2cos(2x)+2+2cos(2x)}=\sqrt{2+2}=\sqrt{4}= 2$

and we see that the second driver is driving faster.

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Calculus 1 : Spatial Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #3 : Derivatives

Example Question #21 : How To Find Velocity

Example Question #22 : How To Find Velocity

Example Question #23 : How To Find Velocity

Example Question #24 : How To Find Velocity

Example Question #25 : How To Find Velocity

Example Question #26 : How To Find Velocity

Example Question #27 : How To Find Velocity

Example Question #28 : How To Find Velocity

Example Question #29 : How To Find Velocity

All Calculus 1 Resources

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