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Calculus 3 : Calculus Review

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #6 : How To Find Velocity

The position of a particle is given by $s(t)=\frac{2t^2+1}{t+1}$ . Find the velocity at t=2 .

Possible Answers:

$\frac{15}{3}$

$\frac{15}{9}$

$\frac{16}{3}$

Correct answer:

$\frac{15}{9}$

Explanation:

The velocity is given as the derivative of the position function, or

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

We can use the quotient rule to find the derivative of the position function and then evaluate that at t=2 . The quotient rule states that

$\left[\frac{f(x)}{g(x)}\right]'=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$ .

In this case, $f(t)=2t^2+1\Rightarrow f'(t)=4t$ and $g(t)=t+1\Rightarrow g'(t)=1$ .

We can now substitute these values in to get

$\left[\frac{f(t)}{g(t)}\right]'=\frac{(t+1)*(4t)-(2t^2+1)*(1)}{(t+1)^2}=\frac{2t^2+4t-1}{(t+1)^2}$ .

Evalusting this at t=2 gives us $\frac{2*2^2+4*2-1}{(2+1)^2}=\frac{16-1}{3^2}=\frac{15}{9}$ .

So the answer is $\frac{15}{9}$ .

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Example Question #1 : Derivatives

The position of an object is given by the following equation:

$s(t)=\frac{2}{3}t^3+\frac{5}{2}t^2-13t+27$

Determine the equation for the velocity of the object.

Possible Answers:

$v(t)=\frac{5}{2}t^2-13t+27$

v(t)=6t^2+2t-13

$v(t)=\frac{2}{3}t^3+\frac{5}{2}t^2-13t$

$v(t)=\frac{2}{3}t^2+\frac{5}{2}t+27$

v(t)=2t^2+5t-13

Correct answer:

v(t)=2t^2+5t-13

Explanation:

Velocity is the derivative of position, so in order to find the equation for the velocity of an object, all we must do is take the derivative of the equation for its position:

$s(t)=\frac{2}{3}t^3+\frac{5}{2}t^2-13t+27$

We will use the power rule to get the derivative.

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

Therefore we get,

v(t)=s'(t)=2t^2+5t-13

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Example Question #2 : Derivatives

The position of a particle is represented by $f(x) = e^{x}+3x^2$ . What is the velocity at x=4 ?

Possible Answers:

Correct answer:

Explanation:

Differentiate the position equation, to get the velocity equation

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

$v(x) = e^x +3\cdot2 x^{2-1}=e^{x} + 6x$

Now we plug 4 into the equation to find the velocity

$v(4) = e^{(4)}+6(4)$

is approximately equal to 2.72. Therefore

$e^{(4)} + 6(4) = 2.72^{(4)} +24 \approx 55+24= 79$

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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 #41 : Calculus

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 velocity of the oxygen molecule.

Possible Answers:

v(t)=84t-28

v(t)=-28t+14

v(t)=42t^2-28t+14

v(t)=42t^2+14

v(t)=14

Correct answer:

v(t)=42t^2-28t+14

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

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Example Question #4 : Derivatives

The position of an object is given by the equation x(t) = e^t +t . What is the velocity of the object at t = 2 ?

Possible Answers:

15.5

9.3

8.3

Correct answer:

8.3

Explanation:

The velocity of the object can be found by differentiating the position equation of the object. To differentiate the position equation of the object, we can use the power rule for the second term where if

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

Using this rule we find that

$v(t) = x'(t) = e^t + (1)t^{(1-1)} = e^t +1$

We can now use the value of t = 2 to solve for the velocity at t = 2

$v(2) = e^2 +1 \approx 2.7^2 +1 = 8.3$

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Example Question #5 : Derivatives

The position of an object is given by the equation $x(t) = 7t^2\sin(t)$ . What is the equation for the velocity of the object?

Possible Answers:

$v(t) = 14t\sin^2(t) - 7t\cos(t)$

$v(t) = 14t\cos(t)$

$v(t) = 14t\sin(t) - 7t^2\cos(t)$

$v(t) = 14t\sin(t) +7t^2\cos(t)$

Correct answer:

$v(t) = 14t\sin(t) +7t^2\cos(t)$

Explanation:

The velocity of the object can be found by differentiating the position equation. The position equation can be accurately differentiated using the power rule and the product rule where if

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

and where if

$f(x)= j(x)k(x) \rightarrow f'(x) = j'(x)k(x) + j(x)k'(x)$

Using these two rules we find the velocity equation to be

$v(t) = x'(t) = 7(2)t^{(2-1)}(\sin(t)) + 7t^2(\cos(t)) = 14t\sin(t) +7t^2\cos(t)$

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Example Question #1 : How To Find Acceleration

The position of a particle is given by $s(t)=e^{t^2}+t^2$ . Find the acceleration of the particle when t=1 .

Possible Answers:

2(3e+1)

2(2e+1)

2(2e-1)

Correct answer:

2(3e+1)

Explanation:

The acceleration of a particle is given by the second derivative of the position function. We are given the position function as

$s(t)=e^{t^2}+t^2$ .

The first derivative (the velocity) is given as

$\frac{d(e^{t^2}+t^2)}{dt}=2te^{x^2}+2t$ .

The second derivative (the acceleration) is the derivative of the velocity function. This is given as

$\frac{d(2te^{t^2}+2t)}{dt}=2e^{t^2}+4t^2e^{t^2}+2$ .

Evaluating this at t=1 gives us the answer. Doing this we get

$a(1)=2e^{1^2}+4*1^2*e^{1^2}+2=6e+2=2(3e+1)$ .

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Example Question #6 : Derivatives

The position of an object is described by the following equation:

$s(t)=\frac{5}{4}t^3-3t^2+\frac{1}{2}t-6$

Find the acceleration of the object at t=1 second.

Possible Answers:

m/s^2

$\frac{5}{4}$ m/s^2

m/s^2

$\frac{15}{2}$ m/s^2

$\frac{3}{2}$ m/s^2

Correct answer:

$\frac{3}{2}$ m/s^2

Explanation:

Acceleration is the second derivative of position, so we must first find the second derivative of the equation for position:

$s(t)=\frac{5}{4}t^3-3t^2+\frac{1}{2}t-6$

$v(t)=s'(t)=\frac{15}{4}t^2-6t+\frac{1}{2}$

$a(t)=v'(t)=s''(t)=\frac{15}{2}t-6$

Now we can plug in t=1 to find the acceleration of the object after 1 second:

$a(1)=\frac{15}{2}(1)-6=\frac{3}{2}$

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Example Question #7 : Derivatives

If h(t) models the distance of a projectile as a function of time, find the acceleration of the projectile at t=6 .

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

Possible Answers:

$51\frac{m}{s^2}$

$165\frac{m}{s^2}$

$48\frac{m}{s^2}$

$15\frac{m}{s^2}$

Correct answer:

$51\frac{m}{s^2}$

Explanation:

We are given a function dealing with distance and asked to find an acceleration. recall that velocity is the first derivative of position and acceleration is the derivative of velocity. Find the second derivative of h(t) and evaluate at t=6.

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

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

h''(t)=8t+3

h''(6)=8*6+3=48+3=51

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Calculus 3 : Calculus Review

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #6 : How To Find Velocity

Example Question #1 : Derivatives

Example Question #2 : Derivatives

Example Question #3 : Derivatives

Example Question #41 : Calculus

Example Question #4 : Derivatives

Example Question #5 : Derivatives

Example Question #1 : How To Find Acceleration

Example Question #6 : Derivatives

Example Question #7 : Derivatives

All Calculus 3 Resources

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