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Calculus 3 : Derivatives

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Example Questions

Example Question #11 : Derivatives

Function v(t) gives the velocity of a particle as a function of time.

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

Find the equation that models that particle's acceleration over time.

Possible Answers:

$\small \small \small \small \small \small \small \small a(t)=5t^4+4t^2-\frac{14t}{3}$

$\small \small \small \small \small \small \small a(t)=20t^3+36t^2-\frac{14}{3}$

$\small \small \small \small \small \small \small a(t)=5t^5+4t^3-\frac{5t}{3}$

$\small \small \small \small \small \small \small a(t)=5t^4+12t^2-\frac{14}{3}$

$\small \small \small \small \small \small a(t)=5t^4+12t^3-\frac{14t}{3}$

Correct answer:

$\small \small \small \small \small \small a(t)=5t^4+12t^3-\frac{14t}{3}$

Explanation:

Recall that velocity is the first derivative of position, and acceleration is the second derivative of position. We begin with velocity, so we need to integrate to find position and derive to find acceleration.

To derive a polynomial, simply decrease each exponent by one and bring the original number down in front to multiply.

So this

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

Becomes:

$\small \small \small \small \small v'(t)=5t^4+12t^3-\frac{14t}{3}$

So our acceleration is given by

$\small \small \small \small \small \small a(t)=5t^4+12t^3-\frac{14t}{3}$

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Example Question #201 : Calculus 3

Consider the following position function:

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

Find the acceleration after seconds of a particle whose position is given by {}p(t) .

Possible Answers:

$33 \frac{m}{s^2}$

$3318 \frac{m}{s^2}$

$318 \frac{m}{s^2}$

$3339 \frac{m}{s^2}$

$33.18 \frac{m}{s^2}$

Correct answer:

$3318 \frac{m}{s^2}$

Explanation:

Recall that acceleration is the second derivative of position, so we need p''(7).

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

Taking the first derivative we get:

p''(t)=72t^2-30t

Taking the second derivative and plugging in 7 we get:

$p''(7)=72\cdot7^2-30\cdot7=3318$

So our acceleration after 7 seconds is

$3318 \frac{m}{s^2}$ .

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Example Question #11 : Derivatives

Given the vector position:

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

Find the expression of the velocity.

Possible Answers:

$\vec{v}=(cos(t)-tsin(t),1,sin(t)+tcos(t))$

$\vec{v}=(cos(t)-tsin(t),1,sin(t)-tcos(t))$

$\vec{v}=(cos(t)-tcos(t),1,sin(t)+tcos(t))$

$\vec{v}=(cos(t)-tsin(t),1,cos(t)+tcos(t))$

$\vec{v}=(cos(t)+tsin(t),1,sin(t)+tcos(t))$

Correct answer:

$\vec{v}=(cos(t)-tsin(t),1,sin(t)+tcos(t))$

Explanation:

All we need to do to find the components of the velocity is to differentiate the components of the position vector with respect to time.

We have :

(tcos(t))'=cos(t)-tsin(t)

t'=1

(tsin(t))'=sin(t)+tcos(t)

Collecting the components we have :

$\vec{v}=(cos(t)-tsin(t),1,sin(t)+tcos(t))$

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Example Question #32 : How To Find Acceleration

A car is driving north on a highway at a constant velocity of mph. What is the acceleration after an hour?

Possible Answers:

Correct answer:

Explanation:

If a car is travelling north at constant velocity 60 mph, it's possible to write a velocity function for this vehicle, where is time in hours.

v(t)=60

To find the acceleration, take the derivative of the velocity function.

a(t)=v'(t)=0

The acceleration after an hour, or any time , is zero.

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Example Question #31 : How To Find Acceleration

Consider the position function {}p(t) , which describes the positon of an oxygen molecule.

p(t)=14t^3-14t^2+14t-14

Find the function which models the acceleration of the particle.

Possible Answers:

a(t)=84t

a(t)=-28

a(t)=84t-28

a(t)=84

a(t)=84-28t

Correct answer:

a(t)=84t-28

Explanation:

Recall that velocity is the first derivative of position and acceleration is the second derivative of position.

So given:

p(t)=14t^3-14t^2+14t-14

Apply the power rule to each term to find the velocity.

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

v(t)=p'(t)=42t^2-28t+14

Applying the power rule a second time we arrive at the acceleration function.

a(t)=p''(t)=84t-28

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Example Question #11 : Calculus Review

The displacement of an object at time is defined by the equation $f(t) = \frac{100}{t}$ . What is the acceleration equation for this object?

Possible Answers:

$\frac{-1}{t^3}$

$\frac{-100}{t^3}$

$\frac{100}{t^3}$

$\frac{200}{t^3}$

Correct answer:

$\frac{200}{t^3}$

Explanation:

The acceleration equation is the second derivative of the displacement equation.

$f(t) = \frac{100}{t} = 100t^{-1}$

Therefore the first derivative is equal to

$f'(t) = (-1)(100)t^{-1-1} = -100t^{-2}$

Differentiating a second time gives

$f''(t) = (-2)(-100)t^{-2-1} = 200t^{-3} = \frac{200}{t^3}$

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Example Question #201 : Calculus 3

Given the velocity function v(t)=3tan(t)sin(t) , find the acceleration function a(t) .

Possible Answers:

a(t)=3cos(t)sec^2(t)

a(t)=3(sin(t)+sec(t)tan(t))

a(t)=3sin(t)+sec(t)tan(t)

a(t)=3(sec^2(t)sin(t)+sin(t)tan(t))

Correct answer:

a(t)=3(sin(t)+sec(t)tan(t))

Explanation:

We know that acceleration a(t) is the derivative of velocity with respect to time.

a(t)=v'(t)

We also know that the velocity function v(t) is given by

v(t)=3tan(t)sin(t)

We need to apply the product rule to solve for the derivative.

Recall that the product rule is given by:

$\frac{\mathrm{d} }{\mathrm{d} t}(f(t)*g(t))=f(t)*g'(t)+g(t)*f'(t)$

In our case, f(t)=tan(t) and g(t)=sin(t)

Therefore,

f'(t)=sec^2(t)

g'(t)=cos(t)

v'(t)=a(t)=3(tan(t)cos(t)+sec^2(t)sin(t))

We can reduce some terms in the acceleration function.

$tan(t)*cos(t)=\frac{sin(t)}{cos(t)}*cos(t)=sin(t)$

$sec^2(t)*sin(t)=\frac{sin(t)}{cos(t)}*sec(t)=tan(t)*sec(t)$

The final answer can be given as

a(t)=3(sin(t)+sec(t)tan(t))

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Example Question #12 : Derivatives

Find the acceteration at , given the velocity function below.

v(t)=2t^2+6t-2

Possible Answers:

Correct answer:

Explanation:

To solve, simply differentiate to find the acceleration function and then plug in .

a(t)=v'(t)=4x+6

a(2)=4(2)+6=14

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Example Question #233 : How To Find Acceleration

The position of a particle traveling along the axis is described by

p(t)=t^3-2t^2-6t-9 .

What is the acceleration rate of the particle when $4\ seconds$ have passed?

Possible Answers:

$26\ \frac{in}{sec^2}$

$28\ \frac{in}{sec^2}$

$-28\ \frac{in}{sec^2}$

$20\ \frac{in}{sec^2}$

$-20\ \frac{in}{sec^2}$

Correct answer:

$20\ \frac{in}{sec^2}$

Explanation:

To find the acceleration of a particle at a specific time we will need to take the derivative of the poistion function twice. Taking the derivative of position onces gives us the velocity function. Taking the derivative a second time will result in the acceleration function.

Since p(t)=t^3-2t^2-6t-9 ,

we will apply the power rule for differentiation which state, $(x^n)'=nx^{n-1}$ .

Therefore applying it once we get,

v(t)=p'(t)=(t^3-2t^2-6t-9)'

$v(t)=3t^{3-1}-2(2t^{2-1})-6t^{1-1}-0$

v(t)=3t^2-4t-6

From here we will apply the power rule one more time to find the acceleration function.

$\\a(t)=v'(t)=(3t^2-4t-6)'\\ \\a(t)=(2*3)t^{2-1}-(1*4)t^{1-1}-0\\ \\a(t)=6t-4$

Now substitute x=4 to find the specific acceleration the question is asking for.

$a(4)=6(4)-4=24-4=20\ \frac{in}{sec^2}$

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Example Question #241 : Acceleration

Find the acceleration at t=1 given the following velocity equation.

v(t)=2t^3-8t^2+14t-2

Possible Answers:

Correct answer:

Explanation:

To solve, simply differentiate v(t) using the power rule to find a(t) . Then plug in for .

Recall the power rule:

f(x)=x^n

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

Apply this to get

a(t)=v'(t)=6t^2-16t+14

a(1)=6(1)^2-16(1)+14=4

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Calculus 3 : Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #11 : Derivatives

Example Question #201 : Calculus 3

Example Question #11 : Derivatives

Example Question #32 : How To Find Acceleration

Example Question #31 : How To Find Acceleration

Example Question #11 : Calculus Review

Example Question #201 : Calculus 3

Example Question #12 : Derivatives

Example Question #233 : How To Find Acceleration

Example Question #241 : Acceleration

All Calculus 3 Resources

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