## Solutions to the 2003 AP Calculus AB Exam (Form B)

### •Problem 1.

#### •a.

If , and , then the curves have an intersection in the first quadrant where .

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We have , so the equation of the line tangent to the curve at is ,
or .  This is , or .

#### •b.

The curve intersects the -axis at , and the line crosses the -axis at .  Therefore the area of the region is

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#### •c.

The curve crosses the -axis at and at .  Thus, the volume generated when the region is revolved about the -axis is

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(Note:  I mistakenly found the area generated by the region instead of the region in an earlier version of these solutions.  Thanks to Heather Kelly for alerting me to the mistake.)

### •Problem 2

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#### •a.

The amount pumped into the tank during is given by .  Integrating numerically, we obtain

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#### •b.

The rate of change of the volume of oil in the tank at time is .  At this is

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The level of oil in the tank is decreasing when because the volume of oil in the tank has negative rate of change then.

#### •c.

We are given that there were gallons of oil in the tank when .  Thus, the volume in the tank at time is, by the Fundamental Theorem of Calculus, .  Integrating numerically with , we obtain

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#### •d.

Here is a plot of the rate at which the volume of oil in the tank changes.

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This rate is positive, and the volume increases, from until about . The rate is negative, and the volume decreases, from about until a little after   Thereafter, volume increases.  This means that there is a minimum somewhere near   Solving numerically, we locate the relevant critical point:

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The amount of oil in the tank is minimal when .

### •Problem 3.

#### •b.

The required midpoint Riemann sum is

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#### •c.

The integral gives the volume, in cubic millimeters, of the segment of the blood vessel that extends from mm to mm.

#### •d.

We have .  Consequently, by Rolle's Theorem, there is a number where .  We also have , so, once again by Rolle's Theorem, there is a number where .  But then , so, yet again by Rolle's Theorem, there must be a number between and , such that .  Noting that , we see that .

### •Problem 4

#### •a.

If , then acceleration is .  When , we have .

#### •b.

Speed is , and so .  When , this is ), which is positive.  Hence is increasing when , because its derivative is positive there.

#### •c.

The particle changes direction where derivative of position with respect to time, which is velocity, changes sign.  This happens only when , or when .

#### •d.

The total distance traveled (as opposed to the total displacement) by the particle over the interval is .  Taking into account the sign change at , this is .  This reduces to .

### •Problem 5

#### •a.

is the area of a rectangle of base 1 and height 2 plus the area of a triangle of base 1 and height 2, or .   By the Fundamental Theorem of Calculus, , and when , this is .  In the interval , the FTC tells us that .  Thus in that interval, and .

#### •b.

The average rate of change of on is .  (See part a for the calculation of ; for , we have , which is the negative of the area of a triangle of base 2 and height 4, or )

#### •c.

By the Fundamental Theorem of Calculus, .  Thus, on the interval , takes on its average value, at two different points:  , and

#### •d.

An inflection point occurs where the monotonicity of the derivative changes from increasing to decreasing or vice versa.  There are two such points for , (We know by the Fundamental Theorem of Calculus.)  They are and .

### •Problem 6

#### •a.

If , then .  Thus,

#### •b.

We have , so, .  Substituting in the integral on the left, we transform it to .  Thus , or .  So .

Converted by Mathematica  (May 14, 2003)