Calculus Review - AP Calculus BC

Card 0 of 835

Question

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

Answer

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 gives us the answer. Doing this we get

$a(1)=2e^{1^2}+41^2e^{1^2}+2=6e+2=2(3e+1)$ .

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Question

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

Answer

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 . 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 gives us $\frac{22^2+42-1}{(2+1)^2}=\frac{16-1}{3^2}=\frac{15}{9}$ .

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

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Question

$f (x) = 7x^{2} + 6x - 9$

Give .

Answer

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f (x) = 7x^{2} + 6x - 9$

$f ' (x) = 7\cdot 2\cdot x^{2-1} + 6\cdot 1 + 0$

$f ' (x) = 14\cdot x^{1} + 6$

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Question

Differentiate $x^{999}$ .

Answer

$\frac{\mathrm{d} }{\mathrm{d} x} \left (x^{n} \right )= n\cdot x^{n-1}$ , so

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{999-1} = 999x^{998}$

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Question

$f(x)= 8x^{3}-7x^{2}+11$

Give .

Answer

First, find the derivative of $f(x)= 8x^{3}-7x^{2}+11$ .

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f(x)= 8x^{3}-7x^{2}+11$

$f'(x)= 3 \cdot 8x^{3-1}-2\cdot 7x^{2-1}+0$

$f'(x)= 24x^{2}-14x^{1}$

$f'(x)= 24x^{2}-14x$

Now, differentiate to get .

$f''(x)= 2 \cdot 24x^{2-1}-1\cdot 14x^{1-1}$

$f''(x)= 48x^{1}-14x^{0}$

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Question

$f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$

Give .

Answer

First, find the derivative of $f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$ .

Recall that $\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0.

$f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$

$f'(x) =4\cdot 0.9x^{4-1} + 3\cdot 0.7x^{3-1}- 1 \cdot 0.5 x ^{1-1}$

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5 x ^{0}$

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

Now, differentiate to get .

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

$f''(x) =3 \cdot 3.6x^{3-1} + 2 \cdot 2.1x^{2-1}- 0$

$f''(x) =10.8x^{2} + 4.2x^{1}- 0$

$f''(x) =10.8x^{2} + 4.2x$

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Question

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.

Answer

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,

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Question

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 second.

Answer

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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Question

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

Answer

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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Question

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

.

Answer

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

So lets take the derivative of with respect to .

The derivative of 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...

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Question

If models the distance of a projectile as a function of time, find the acceleration of the projectile at .

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

Answer

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.

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

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Question

Function 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.

Answer

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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Question

Consider the following position function:

Find the acceleration after seconds of a particle whose position is given by .

Answer

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

Taking the first derivative we get:

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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Question

Given the vector position:

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

Find the expression of the velocity.

Answer

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 :

Collecting the components we have :

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

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Question

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

Answer

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.

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

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

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Question

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?

Answer

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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Question

Consider the position function , which describes the positon of an oxygen molecule.

Find the function which models the velocity of the oxygen molecule.

Answer

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

So given:

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

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

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Question

Consider the position function , which describes the positon of an oxygen molecule.

Find the function which models the acceleration of the particle.

Answer

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

So given:

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

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

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

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Question

The position of an object is given by the equation . What is the velocity of the object at ?

Answer

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 to solve for the velocity at

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

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Question

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?

Answer

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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