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AP Calculus AB : AP Calculus AB

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Example Questions

Example Question #611 : Ap Calculus Ab

A ball is thrown into the air. It's height, after t seconds is modeled by the formula:

h(t)=-15t^2+30t feet.

At what time will the velocity equal zero?

Possible Answers:

1.5s

Correct answer:

Explanation:

In order to find where the velocity is equal to zero, take the derivative of the function and set it equal to zero.

h(t) = –15t²+ 30t

h'(t) = –30t + 30

0 = –30t + 30

Then solve for "t".

–30 = –30t

t = 1

The velocity will be 0 at 1 second.

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Example Question #612 : Ap Calculus Ab

If oil is leaking from a pipe at a rate of gallons per hour, in which is the time in hours, how many gallons will have leaked out hours after the leaking has begun?

Possible Answers:

gallons

Correct answer:

gallons

Explanation:

For this problem we need to recognize that we will need to take the integral of the rate over the specific time period the question is asking about.

Our function is,

$\text{ Rate} = (2t) \text{ gallons}/ \text{ hour}$ .

Therefore to see how many gallons will leak out 3 hour later you take the integral of 2t and evaluate it from t=0 to t=3.

$\int_{0}^{3} 2t= \frac{2t^2}{2}=t^2$

=(3)^2-(0)^2

Therefore you find that 9 gallons leak out during the three hour interval.

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Example Question #613 : Ap Calculus Ab

If water drains from a tank at a rate of 3t^2 gallons an hour, where is the time in hours, how much water will have drained hours after the initial leak?

Possible Answers:

gallons

gallon

gallons

Correct answer:

gallons

Explanation:

For this problem we need to recognize that to find the total amount of water drained over a certain time interval at a specific rate, we need to take the integral of the rate and evaluate it on the specified interval.

$\text{Rate} = 3t^2 \text{gallons}/ \text{hour}$

$\int_{0}^{2}3t^2 dt = (t^3)_{0}^{2}$

=(2)^3-(0)^3

After two hours 8 gallons of water will have leaked.

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Example Question #614 : Ap Calculus Ab

True or False: If f(x) is a negative-valued function for all , $\int_0^1 f(x) dx <0$ .

Possible Answers:

True

False

Correct answer:

True

Explanation:

This is true. Since f(x) is negative-valued, its graph is below the -axis, and the Riemann sums used to evaluate the area between f(x) and the -axis have a negative value for height.

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Example Question #1 : Definite Integral As A Limit Of Riemann Sums

f(x) is a continuous function on the interval $\left [ 1,10 \right ]$ and is differentiable on the open interval $\left ( 1,10 \right )$ . If f(1)=f(10) , then which of the following statements MUST be true:

Possible Answers:

$f^{'}(x)=0$ over the interval $\left [ 1,10 \right ]$ .

$f^{'}(x)>0$ over the interval $\left [ 1,10 \right ]$ .

$f^{'}(x)<0$ over the interval $\left [ 1,10 \right ]$ .

$f^{''}(c)=0$ at some point , where 1<c<10 .

$f^{'}(c)=0$ at some point , where 1<c<10 .

Correct answer:

$f^{'}(c)=0$ at some point , where 1<c<10 .

Explanation:

According to Rolle's Theorem, if a function is continuous over a closed interval $\left [ a,b \right ]$ and differentiable on the open interval $\left ( a,b \right )$ , and if f(a)=f(b) , then there has to be some value such that $f^{'}(c)=0$ , where a<c<b . To put this another way, if a function is continuous and differentiable over a certain interval, and if the value of the function is the same at both endpoints of that interval, then at some point in between those endpoints, the function is going to have a slope of zero (i.e. its first derivative will be zero). This does not apply to the second derivative, nor does it require that the slope of the first derivative be zero over the entire interval.

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Example Question #1 : Definite Integral As A Limit Of Riemann Sums

You may use one or both of the following summation formulas:

$\sum_{i=1}^n i = \frac{n(n+1)}{2}$

$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$

Express the following definite integral as a limit of Riemman Sums. Then evaluate the integral by evaluating the limit.

$\int_1^2 (x-1)^2dx$

Possible Answers:

$\frac{8}{3}$

$\infty$

$\frac{1}{3}$

$\frac{16}{3}$

Correct answer:

$\frac{1}{3}$

Explanation:

First, let's remind ourselves what the formula for Riemann Sums is. Written in Summation form, it is:

$\sum_{i=1}^n f(x_i)\Delta x$

when , the number of subintervals, is small, the area calculated by the Riemann Sums is not very accurate. However, as we increase the number of subintervals, the approximation becomes closer and closer to the exact area under the function. "Closer and closer" is a concept from Limits. So if we find the limit of the Riemann sum formula, with n approaching infinity, the result is the exact area. This is the essence of the Definite integral definition. What it effectively tells us to do is stick a limit on the Riemann sums formula to get:

$Exact Area=\lim_{n \to \infty} \sum_{i=1}^n f(x_i)\Delta x$

To use this formula, we need to do three things:

(1) we need to find $\Delta x$

(2) we need to develop a formula for x_i

(3) we need to plug that x_i into the function inside the given integral.

To find $\Delta x$ , we use the same formula from Riemann sums.

$\Delta x = \frac{b-a}{n}$ , where and are the upper and lower bounds from the definite integral. For our integral, a=1 and b=2 . Since we are not stating a specific number of subintervals, we leave as it is. Nothing gets plugged into it.

Plugging in a=1 and b=2 , we get

$\Delta x = \frac{2-1}{n}=\frac{1}{n}$

Now we have $\Delta x$ . Next we find a formula for x_i , which by the way represents the right endpoint of the "i-th" subinterval. We will revisit that sentence shortly to clarify.

First, we know that the left endpoint of the 1st subinterval is a = 1 , and that this subinterval has a width of $\Delta x = \frac{1}{n}$ . So to get the right endpoint of the 1st subinterval, we just add that width to a=1

$x_1 = 1 + 1(\Delta x)$

To get the 2nd right endpoint, we just add a 2nd $\Delta x$ .

$x_2 = 1 + 2(\Delta x)$

to get the "i-th" right endpoint, we just add $\Delta x$ "i" times.

$x_i = 1 + i \cdot\Delta x$

This is the "i-th" concept. Now if we wanted the 50th rigth endpoint, we would just plug 50 in for "i". Fortunately, we don't need to.

Now we have $\Delta x$ and we have a formula for x_i . Next we will find f(x_i) by plugging our formula in to the function from inside our integral. In other words our function is f(x)=(x-1)^2 .

Plugging in to this function we get

f(x_i)= (x_i -1)^2

$f(x_i)= ((1+i\cdot\Delta x)-1)^2$

Recall that we determined $\Delta x = \frac{1}{n}$ . Pluggin that in, we get

$f(x_i)= (1+i(\frac{1}{n})-1)^2$

To simplify, we will combine the 1 and -1 to cancel them, and then apply the square power to what is left.

$f(x_i)=(\frac{i}{n})^2$

$f(x_i)=\frac{i^2}{n^2}$

Now we have all the pieces of the Limit of Riemann sums formula. Let's plug them in.

$\lim_{n \to \infty} \sum_{i=1}^n f(x_i)\Delta x$

$\lim_{n \to \infty} \sum_{i=1}^n [(\frac{i^2}{n^2})(\frac{1}{n})]$

To evaluate the limit, we must first find the sum. To do this, we must realize that we can pull any common factors out of the summation, as long as they are constant with respect to the summation index . In other words, we can pull out any variable that is not an . This means we can write the 's on the left of the sum, while still writing them inside the limit,

like this:

$\lim_{n \to \infty} \frac{1}{n^3}\sum_{i=1}^n i^2$

By passing the 's to the left of the sum, we have matched our sum to one of the provided sum formulas, specifically:

$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$

This means we can evaluate the sum by replacing it using the above formula.

Doing this, we get:

$\lim_{n \to \infty} \frac{1}{n^3} [\frac{n(n+1)(2n+1)}{6}]$

Now we can start evaluating the limit.

Lets start by reducing the single from top and with one from the n^3 on bottom.

$\lim_{n \to \infty} \frac{1}{n^2} [\frac{(n+1)(2n+1)}{6}]$

Now we will multiply our the two factors on top, and also multiply the $\frac{1}{n^2}$ through the other fraction.

$\lim_{n \to \infty} \frac{2n^2 +n +2n +1}{6n^2}$

Combining like terms, we get

$\lim_{n \to \infty} \frac{2n^2 +3n +1}{6n^2}$

Since the denominator is a single term, we can split the big fraction into three individual fractions, and reduce each one.

$\lim_{n \to \infty} \frac{2n^2}{6n^2}+ \frac{3n}{6n^2} + \frac{1}{6n^2}$

$\lim_{n \to \infty} \frac{1}{3}+ \frac{1}{2n} + \frac{1}{6n^2}$

Remember that the limit of a constant is equal to that constant. Also remember when evaluating limits at infinity, a constant divided by a variable equals zero.

Applying these, we get

$\frac{1}{3}+ 0 + 0$

This gives us the correct answer of

$\frac{1}{3}$

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Example Question #615 : Ap Calculus Ab

Find two numbers whose sum is 32 and the product is as large as possible.

Possible Answers:

12,20

13,19

16,16

15,17

14,18

Correct answer:

16,16

Explanation:

If one number is , the other number is 32-x . The function that describes this is y=x(32-x) .

The derivative of this is 32-2x .

Setting that derivative equal to 0 and solving for , yields the correct answer of 16.

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Example Question #616 : Ap Calculus Ab

Suppose that a firm's marginal revenue can be modeled by the equation r=10x and its marginal costs can be modeled by $x^3-6x^2+15x$ . What is the maximum profit that firm can achieve?

Possible Answers:

$x=2\pm\frac{\sqrt{21}}{3}$

$x=2\pm\sqrt\frac{17}{3}$

$x=-2\pm\sqrt\frac{7}{3}$

$x=-2\pm\sqrt\frac{17}{3}$

Correct answer:

$x=2\pm\frac{\sqrt{21}}{3}$

Explanation:

If r=10x then marginal revenue = 10

If c=x^3-6x^2+15x then marginal costs = 3x^2-12x+15

Maximum profit occurs when marginal cost = marginal revenue.

Setting the two equations equal and solving for yields the correct solution.

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Example Question #12 : Derivative At A Point

Find the minimum value of $f(x)=\frac{1}{\sqrt{9-x^2}}$

Possible Answers:

$-\frac{1}{3}$

$\frac{1}{3}$

Correct answer:

$\frac{1}{3}$

Explanation:

In order to find the extreme value we need to take the derivative of the function.

$f'(x)=\frac{x}{(9-x^2)^{\frac{3}{2}}}$

After setting it equal to 0, we see that the only candidate is for x=0 . After setting x=0 into f(x) , we get the coordinate $(0,\frac{1}{3})$ as an extreme value. To confirm it is a minimum we can plot the function.

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Example Question #1 : How To Find Maximum Values

What is the local maximum of x^3+2x^2+5 when $x\leq 0$ ?

Possible Answers:

There is no local maximum.

$x=-\frac{2}{3}$

$x=-\frac{1}{3}$

x=0

$x=-1\tfrac{1}{3}$

Correct answer:

$x=-1\tfrac{1}{3}$

Explanation:

To find the maximum, we need to look at the first derivative.

To find the first derivative, we can use the power rule. To do that, we lower the exponent on the variables by one and multiply by the original exponent.

We're going to treat as 5x^0 since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3*x^{3-1})+(2*2x^{2-1})+(0*5x^{0-1})$

Notice that $(0*5x^{0-1}) =0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3*x^{3-1})+(2*2x^{2-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3*x^{2})+(2*2x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=3x^2+4x$

When looking at the first derivative, remember that if the output of this equation is positive, the original function is increasing. If the derivative is negative, then the function is decreasing.

Notice that 3x^2+4x changes from positive to negative when x<0 .

We can find that root using the quadratic equation: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{-4 \pm \sqrt{4^2 - 4(2)(0)}}{2(3)}$

$x = \frac{-4 \pm \sqrt{16 - 0}}{6}$

$x = \frac{-4 \pm 4}{6}$

Since we're looking for a negative value, we will subtract.

$x = \frac{-4 - 4}{6}$

$x = \frac{-8}{6}=-1\tfrac{1}{3}$

Therefore, the maximum is at $x=-1\tfrac{1}{3}$ .

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #611 : Ap Calculus Ab

Example Question #612 : Ap Calculus Ab

Example Question #613 : Ap Calculus Ab

Example Question #614 : Ap Calculus Ab

Example Question #1 : Definite Integral As A Limit Of Riemann Sums

Example Question #1 : Definite Integral As A Limit Of Riemann Sums

Example Question #615 : Ap Calculus Ab

Example Question #616 : Ap Calculus Ab

Example Question #12 : Derivative At A Point

Example Question #1 : How To Find Maximum Values

All AP Calculus AB Resources

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