Math 2250 HW #11 Solutions

Math 2250 Written HW #11 Solutions

1. An airplane begins its descent toward the runway when it is 4 miles from the touchdown point and at an altitude of 1 mile.

(a) Find a, b, c, d so that the cubic function f (x) = ax3 + bx2 + cx + d describes a smooth glide path for the airplane as pictured below (Hint: you have 4 pieces of information about the function f (x). If you translate these pieces of information into 4 equations, you will be able to solve for the four unknowns a, b, c, and d.)

4,1 1

4

3

2

1

Answer: Since f (x) to gives the altitude of the plane when it is x miles from the runway, we're given four pieces of information in the statement of the problem:

f (0) = 0 f (-4) = 1

f (0) = 0 f (-4) = 0

The first two just say that the plane goes through the points (0, 0) and (-4, 1). The third says that the plane's trajectory must level off to horizontal when it lands. The last says that the plane was in level flight at the point (-4, 1), which is the point where it started descending. Now, let's use these four pieces of information. From the first equation, we know that f (0) = 0, so

f (0) = a(0)3 + b(0)2 + c(0) + d 0=0+0+0+d 0 = d.

So in fact the function f (x) = ax3 + bx2 + cx for some a, b, and c. Now, the second equation tells us that f (-4) = 1, so

f (-4) = a(-4)3 + b(-4)2 + c(-4) 1 = -64a + 16b - 4c.

1

We'll keep this equation in our back pocket for later use. Now, the last two equations tell us about the derivative of f (x) = ax3 + bx2 + cx, so let's first compute the derivative:

f (x) = 3ax2 + 2bx + c.

Since the third equation says that f (0) = 0, we know that

f (0) = 3a(0)2 + 2b(0) + c 0=0+0+c 0=c

So now we know that c = 0 and so f (x) = ax3 + bx2 and f (x) = 3ax2 + 2bx. Also, we can simplify the equation 1 = -64a + 16b - 4c to read 1 = -64a + 16b. Finally, the fourth equation tells us that

f (-4) = 3a(-4)2 + 2b(-4) 0 = 48a - 8b.

So now we have the system

0 = 48a - 8b 1 = -64a + 16b

which we want to solve for the two unknowns. From the first equation we know that

48a = 8b,

so after dividing both sides by 8 we have that

b = 6a.

We can substitute this into the other equation:

1 = -64a + 16b 1 = -64a + 16(6a) 1 = -64a + 96a 1 = 32a

1 a= .

32

Since

b

=

6a,

this

tells

us

that

b

=

6 32

=

3 16

,

and

we

can

conclude

that

f (x) = 1 x3 + 3 x2 32 16

is the function we're looking for.

2

(b) Assume the plane follows the path you found in part (a). When is the plane descending at the greatest rate?

Answer: The plane is descending at the greatest rate when the function f (x) (which gives the rate of descent) has its absolute minimum given the constraint -4 x 0. Therefore, we should look for critical points of the function

f (x) =

3 x2 +

6 x=

3

x2

+

3 x.

32 16 32 8

As always when looking for critical points, we differentiate:

6 33 3 f (x) = x + = x + .

32 8 16 8

This function always exists, so the only critical points occur when it equals zero:

33 f (x) = x +

16 8 33 0= x+ 16 8 33 -= x 8 16

-2 = x.

The only critical point of f (x) is at x = -2, so we just have to evaluate f (x) at x = -2 and at the endpoints x = -4 and x = 0: the smallest value that results will be the absolute minimum. Of course, we already know that f (0) = 0 and that f (-4) = 0, so we can just compute

f (-2) =

3

(-2)2

+

3 (-2)

=

12

-

6

=

3

-

6

=

3 -.

32

8

32 8 8 8 8

Since this is smaller than 0, we conclude that f (x) achieves its absolute minimum at x = -2. In other words, the plane is descending at the greatest rate when it is 2 miles from the runway, where its altitude is

f (-2) =

1 (-2)3 +

3

(-2)2

=

1 -

+

3

=

1

mile.

32

16

44 2

2. Determine whether the following statements are true or false. If the statement is true, explain why. If it is false, give an example which shows that it is false (called a "counterexample").

(a) The sum of two increasing functions is increasing. Answer: True. Assuming f (x) and g(x) are differentiable functions which are both increasing, then we know that f (x) and g (x) are both positive. But then if h(x) = f (x) + g(x), it must be the case that

h (x) = f (x) + g (x)

is also positive, since the sum of two positive numbers is positive. Therefore, the function h(x) is also increasing.

3

(b) The product of two increasing functions is increasing. Answer: False. Consider the functions f (x) = ex and g(x) = ex - 10. Notice that both f (x) and g(x) have ex as their derivative. Since ex > 0 for all x, this means that f (x) and g(x) are always increasing, as we can see in the following graph:

30

20

10

4

2

10

2

4

Then the product of f (x) and g(x) is the function h(x) = f (x)g(x) = ex(ex - 10) = e2x - 10ex.

But now

h (x) = 2e2x - 10ex,

which is not always positive. Indeed, for x < ln(5), h (x) < 0 (for example, h (0) = -8). Therefore, the function h(x) is not always increasing, as we can see in the graph:

120 100

80 60 40 20

4

2

20

2

4

(In fact, with slightly more work we could have come up with an example where both functions are always increasing but the product is always decreasing. I won't go through the calculations, but consider

f (x) = g(x) = arctan(x) - .

2

Then f (x) and g(x) are the same increasing function as we can see on the graph:

4

2

2

2

but the product f (x)g(x) = (arctan(x) - /2)2 is always decreasing:

2

2

Where do these examples come from? Well, if f (x) and g(x) are increasing and differentiable, then we know that f (x) > 0 and that g (x) > 0. Now, the product h(x) = f (x)g(x) is increasing where its derivative is positive. But the derivative of h(x) is

h (x) = f (x)g(x) + f (x)g (x)

by the product rule. We know f (x) and g (x) are positive, but if either f (x) or g(x) (or both) is negative, then the whole expression can be negative. Then it's just a matter of coming up with such functions.)

3. Use your accumulated calculus skills to sketch the graph of the function

x g(x) = 1 + x2 .

Be sure to label all intercepts, local minima, local maxima, inflection points, asymptotes, absolute minima, absolute maxima, etc.

Answer: First, we can identify intercepts in the usual way. The y-intercept occurs when

x = 0:

0 g(0) = 1 + 02 = 0

so the y-intercept is at (0, 0).

The x-intercepts occur when g(x) = 0, meaning that

x 1 + x2 = 0,

5

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