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

### •Problem 1.

#### •a.

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

In[1]:=

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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

In[2]:=

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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

In[3]:=

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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

#### •a.

The circle of radius 2 has equation in the first quadrant and extends over the interval there. The circle of radius 1 has equation in the first quadrant and extends over the interval there.  Using the fact that the two semi-circles intersect at , the required expression is therefore .

#### •b.

The circle of radius 2 has equation in the first quadrant and extends over the interval there.  The first-quadrant portion of the circle of radius 1 that lies inside the larger circle has equation and extends over the interval .  The required expression is therefore .

#### •c.

Note:  A phrase seems to have been left out of the distributed version of this question.  I assume that the blank space between the symbol " "  and the symbol  " " should contain the phrase "that represents the area of", or some phrase of similar meaning.

The segment connecting the origin to the first-quadrant intersection point of the two circles, given as lies in the ray .  The area of the sector of the large circle that lies below this segment but above the positive -axis is a part of and has area .  The remainder of the region after this sector is removed has bounding curve , where .  Hence, its area is .  The required expression is therefore .

### •Problem 3.

#### •b.

The required midpoint Riemann sum is

In[4]:=

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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.

The velocity vector is .  At  , this is , and this vector has length , which is speed at .

as .

#### •c.

The tangent line is horizontal when .  By part b, above, the tangent line can be horizontal only when .   Because both exponentials are always positive, there are no horizontal tangent lines.

#### •d.

The tangent line is vertical precisely when the tangent  vector is vertical.  By part a, the tangent line can be vertical only when .  This is equivalent to , or .  (It should be noted that we saw in part c., above, that never takes on the value zero; if were zero, this would complicate our analysis.)

### •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.

The Taylor series about for has general term , where .  Through the first four terms, we have .  The general term is .

#### •b.

We apply the Ratio Test:  .  This limit is less than when , so the radius of convergence for this series is .

#### •c.

We have , and we may integrate the series term-by-term (as long as ).  Thus, .  The general term is .

#### •d.

When , the general term of the series for becomes .  As , this does not converge to zero, so the series of which it is the -th term cannot converge.

Converted by Mathematica  (May 14, 2003)