NAME:
Calculus with Analytic Geometry I
Exam 5-Take Home Part
Due: Monday, October 3, 2011; 12PM
INSTRUCTIONS. As usual, show work where appropriate. As usual, use equal signs properly, write in full
sentences, and be clear. Try to work on these pages, but if you need more paper, USE MORE PAPER.
These instructions should not be there.They make no sense. They came from the in class exam that I used as
a template.
1. (Very similar, but not quite the same as Exercise 65, §3.4) If f and g are the functions whose graphs are
shown, let u(x) = f (g(x), v(x) = g(f (x)), and w(x) = g(g(x)). Find each derivative if it exists. If it doesn’t
exist, explain why. I need to see some work here for credit!
Solution. The fact that what was required where the values of the derivatives at x = 2. Without that
information the exercise becomes considerably more difficult. Unfortunately, with the information I added
by e-mail, the exercise becomes absurdly easy. The answer is: None of the derivatives exist at x = 2.
If you selected x = 1 (as in the textbook), then
1
= f 0 (g(1))g 0 (1) = f 0 (4)g 0 (1) = (− )(−4) = 1
4
v 0 (1) = g 0 (f (1))f 0 (1) = g 0 (2)f 0 (1) does not exist
2
8
w0 (1) = g 0 (g(1))g 0 (1) = g 0 (4)g 0 (1) = × (−4) = −
3
3
u0 (1)
2. (Similar, but not the same as Exercise 67 of §3.4) If g(x) =
g 0 (2).
p
f (x), where the graph of f is shown, evaluate
2
1
1
dg
= f (x)−1/2 f 0 (x). Setting x = 2, g 0 (2) = f (2)−1/2 f 0 (2). Looking at
dx
2
2
the picture one sees that f (2) = 2. To estimate f 0 (2), there is a part of the picture where we can see that the
tangent line goes down four units as the variable x moves three units to the right. This means that it has
slope −4/3, thus f 0 (2) = −4/3. All in all,
1 −1/2
4
2
0
g (2) = · 2
−
=− √ .
2
3
3 2
Solution.
By the chain rule,
3. (§3.5, # 34)
(a) Find an equation of the tangent line to the curve with equation y 2 = x3 + 3x2 at the point (1, −2).
Solution. Step 1. Finding the slope. By implicit differentiation, 2yy 0 = 3x2 + 6x. Setting
x = 1, y = −2, we get −4y 0 = 9, thus y 0 = −9/4.
Because solving for y is not terribly hard, one can also do this directly; usually this is much harder or
impossible. One has y = ±(x3 + 3x2 )1/2 and one has to decide which sign to choose. Setting x = 2, we
get y = ±2, so we have to choose the negative sign. Now
1 3
9
3
2 1/2 dy 2 −1/2
2
y = −(x + 3x ) ,
= − (x + 3x )
(3x + 6x)
=− .
dx x=1
2
4
x=1
9
noindentStep 2. Finishing the problem. The point slope form of the tangent is y + 2 = − (x − 1).
4
9
1
Equivalently, y = − x + .
4
4
(b) At what points does this curve have horizontal tangents? (A horizontal tangent is a tangent with 0
slope).
3x2 + 6x
Solution. From the work done in part a (the implicit differentiation) we have y 0 =
. The
2y
tangent line is horizontal when y 0 = 0; a fraction is 0 if and only if its numerator is 0 (and the denominator
is not!), so we see we’ll have a horizontal tangent line exactly when 3x2 + 6x = 0. This equation can also
be written in the form 3x(x + 2) = 0; the solutions are x = 0 and x = −2. To find the corresponding y
values, we go back to the original equation y 2 = x3 + 3x2 . For x = 0 we get y = 0. However, we should
be wary of this point; looking at the equation we have for y 0 we also have a 0 in the denominator. The
3
graph will give us more information. Setting x = −2, we get y 2 = 4, thus y = ±2. This gives us two
points for sure with horizontal tangent lines, namely (−2, 2) and (−2, −2).
(c) (Optional) Illustrate parts (a) and (b) by graphing the curve and the tangent lines on a common screen.
And printing out the result and attaching it to this exam.
4. (§3.5, # 40) If x2 + xy + y 3 = 1, find the value of y 000 at the point where x = 1. It should go without saying
that this value has to be a number.
Solution. By implicit differentiation, 2x + y + xy 0 + 3y 2 y 0 = 0. We could solve this for y 0 , but because we
are required to find a specific value, it will be easier to continue differentiating implicitly. Next
d
(2x + y + xy 0 + 3y 2 y 0 ) = 0;
dx
thatis,2 + y 0 + y 0 + xy 00 + 6yy 0 y 0 + 3y 2 y 00 = 0.
We can rewrite this as 2 + 2y 0 + 6y(y 0 )2 + xy 00 + 3y 2 y 00 = 0. Applying once more d/dx, we get
2y 00 + 6(y 0 )3 + 12yy 0 y 00 + y 00 + xy 000 + 6yy 0 y 00 + 3y 2 y 000 = 0.
We write out all our equations, grouping together some terms.
x2 + xy + y 3
=
1
0
=
0
2
2x + y + (x + 3y )y
0
0
2
2y + 6y(y )2 + (x + 3y )y
00
0 3
0
00
2
2y + 6(y ) + (18yy + 1)y + (x + 3y )y
00
000
= −2
=
0
Now set x = 1. The first equation becomes 1 + y + y 3 = 1, thus y(1 + y 2 ) = 0, hence y = 0. In the second
equation set x = 1, y = 0. It becomes 2 + y 0 = 0, hence y 0 = −2. Now set x = 1, y = 0, y 0 = −2 in the
third equation, to get −4 + y 00 = −2, hence y 00 = 2. Finally, setting x = 1, y = 0, y 0 = −2, y 00 = 2 in the last
equation gives 4 − 48 + 2 + y 000 = 0, whence y 000 = 42.
4
Note: So far, these are fairly straightforward exercise. If you do them all right, you get 90points. For 100 points
you have to do at least one of the following exercises. Each one adds 10 points to your total, so your grade here
could be more than 100 points.
√
√
√
5. (§3.5, #46) Show that the sum of the x- and y-intercepts of any tangent line to the curve x + y = c is
equal to c.
Solution. Suppose we have the tangent line at a point (x0 , y0 ) of the curve. By implicit differentiation
√
y
x−1/2
1 −1/2 1 −1/2 0
√ √
x
+ y
y = 0 so that y 0 = − −1/2 = √ . The slope at (x0 , y0 ) is thus − y0 / x0 and the equation
2
2
x
y
of the tangent line is
√
√
y0
y0
√ √
y − y0 = − √ (x − x0 ); solving for y, etc. y = − √ x + x0 y0 + y0 .
x0
x0
√
y0
√ √
√ √
(We used that x0 √ = x0 y0 .) The y-intercept is obtained by setting x = 0 and works out to x0 y0 +y0 .
x0
The x intercept is obtained setting y = 0; we get
√
x0 √ √
√ √
x = √ ( x0 y0 + y0 ) = x0 + x0 y0 ,
y0
√ √
so that the sum of the two intercepts works out to x0 + y0 + 2 x0 y0 . For the final touch, one has to
√
√
recognize that this is a binomial expansion of the form a2 + b2 + 2ab, if a = x0 , b = y0 . That is,
√
√ √
√
√ 2
x0 + y0 + 2 x0 y0 = ( x0 + y0 ) = ( c)2 = c.
Since (x0 , y0 ) was an arbitrary point on the curve, that is all it takes.
6. Exercise #80, Section 3.5 of the textbook:
Solution. One thing that makes this problem look more difficult than it really is, is a typical magical trick:
Your attention is drawn in one direction, things are happening elsewhere. Here your attention is drawn to
the lamp. Forget about the lamp for a moment; it is hardly relevant. The problem could easily have been
phrased as follows: Draw a tangent to the ellipse from (−5, 0). Where does this tangent intersect the line
x = 3? Let us call (x0 , y0 ) the point on the ellipse where this tangent is actually tangent to the ellipse. By
implicit differentiation 2x + 8yy 0 = 0, thus y 0 = −x/4y; at (x0 , y0 ) we get a slope m = −x0 /4y0 . Notice
m > 0 since we are taking y0 > 0 and x0 < 0. The equation of the tangent line is thus y = m(x + 5). But
we still don’t know what m is. But we have two further conditions. One is that the point (x0 , y0 ) is on the
line, second that it is on the ellipse. Being on the line means that y0 = m(x0 + 5) = (−x0 /4y0 )(x0 + 5);
multiplying by 4y0 and rearranging we get x20 + 4y02 = −5x0 . Being on the ellipse means x20 + 4y02 = 5 so that
we must have 5 = −5x0 or x0 = −1. Going back to the equation for the ellipse, we now have 1 + 4y02 = 5,
hence y0 = 2 (being positive). Thus m = 1/4 and now setting x = 3 in y = (1/4)(x + 5), we get the answer
y = 2.
5
7. Problem 8 from the Problems Plus Section of Chapter 3 (p.270). A car is traveling at night along a highway
shaped like a parabola with its vertex at the origin (see the figure). The car starts at a point 100m west
and 100 m north of the origin and travels in an easterly direction. There is a statue located 100m east and
50m north of the origin. At what point on the highway will the car’s headlights illuminate the statue. Your
answer, of course, should be something like at a point x east/west and y meters north/south of the origin.
And it should be clear how you arrived at your answer.
Solution. A lot of information is sort of hidden. For example, the parabola has the vertex at (0, 0) and
has the y-axis as its axis. It stays in the upper half-plane. The only parabolas satisfying all these conditions
have equations y = ax2 , with a > 0. We are also told that the point (−100, 100) is on the parabola; thus
1 2
100 = a(−100)2 = 10, 000a. So a = 1/100. The equation of the parabola is y =
x . Suppose that the
100
car’s headlights illuminate the statue at the point of coordinates (x0 , y0 ). That means that the tangent line of
1 2
the parabola at that point goes through the point (100, 50). We know: y0 =
x . The slope at that point is
100 0
1
1
x0 , so the equation of the tangent line is y − y0 =
x0 (x − x0 ). Finally, that the tangent line goes through
50
50
1
1
(100, 50) tells us that 50 − y0 =
x0 (100 − x0 ). This last equation can be rewritten as 50 − y0 = 2x0 − x20 .
50
50
1 2
Using that y0 =
x , we can simplify this to y0 = 2x0 − 50. But, from the parabola equation, y0 = x20 /100,
100 0
so now we have that (multiplying by 100)
x20 − 200x0 + 5000 = 0.
√
Solving we have two solutions x = 100 ± 50 2. The one with the minus sign is the one we are looking for, the
other one corresponds to when the tailights of the car, if they were powerful enough, would be illuminating
the statue. Squaring and dividing by 100, we get the y0 value.
√
√
The car is (100 − 50 2)m east and 150 − 100 2m north of the origin.
Notice The pictures in the posted version are in color.