Math 2250 Written HW #7 Solutions
1. The following plot shows the graph of a function f (x) along with the graphs of f 0 (x) and
f 00 (x). Which is which? Explain your reasoning.
B
A
C
Answer: Notice that A is negative whenever B is decreasing, and A is positive whenever B
is increasing. Therefore, it’s almost certain that A is the graph of the derivative of B. In turn,
C is negative when A is decreasing and C is positive when A is increasing, so C should be
the graph of the derivative of A. Moreover, the roots of B don’t match up with the locations
of horizontal tangent lines for either A or C, so it shouldn’t be the case that B is the graph
of the derivative of either.
Therefore, we can conclude that B is the graph of f (x), A is the graph of f 0 (x), and C is the
graph of f 00 (x).
2. Let h(t) be the height of a falling ball at time t. If the ball is in freefall, the velocity of the
ball when the ball is at height h is given by
p
v(t) = k h(t)
for some constant k. Show that the acceleration of the ball is constant.
Answer: First of all, by definition acceleration is the rate of change of velocity with respect
to time, so acceleration is given by
a(t) = v 0 (t).
√
Now, we can compute the derivative of v using the chain rule: if we let g(u) = k u, then
v(t) = (g ◦ h)(t), so
k
a(t) = v 0 (t) = g 0 (h(t)) · h0 (t) = p
· h0 (t).
2 h(t)
0
But
p h (t) is the rate of change of height with respect to time, which of course is just v(t) =
k h(t). Therefore,
p
k
k
k2
k
a(t) = p
· h0 (t) = p
· v(t) = p
· k h(t) = ,
2
2 h(t)
2 h(t)
2 h(t)
which is indeed constant.
1
3. Shown below is the curve
x3 + y 3 − 9xy = 0,
which is known as the folium of Descartes. At what point other than the origin does the
curve have a horizontal tangent line?
5
4
3
2
1
-5
-4
-3
-2
-1
1
2
3
4
5
-1
-2
-3
-4
-5
Answer: To find the slope of the tangent line, we can use implicit differentiation:
d
d
x3 + y 3 − 9xy =
(0)
dx
dx
d
d
d
x3 +
y 3 − 9 (xy) = 0
dx dx
dx d
d
d
3 dy
2
y
−9
(x) · y + x ·
(y) = 0
3x +
dy
dx
dx
dx
dy
2
2 dy
3x + 3y ·
−9 y+x·
=0
dx
dx
dy
dy
− 9y − 9x ·
= 0.
3x2 + 3y 2 ·
dx
dx
We can isolate the terms containing
dy
dx
on the left hand side and then solve for
dy
dx :
dy
dy
− 9x
= 9y − 3x2
dx
dx
dy
3y 2 − 9x = 9y − 3x2
dx
dy
9y − 3x2
= 2
.
dx
3y − 9x
3y 2
So the slope of the tangent line to the curve at the point (x, y) is
horizontal tangent line if and only if
9y − 3x2
=0
3y 2 − 9x
2
9y−3x2
.
3y 2 −9x
Hence, we have a
which is to say, if and only if
9y − 3x2 = 0.
2
Solving for y yields that y = x3 . In other words, a point (x, y) on the curve will have a
2
horizontal tangent line if y = x3 , so we need to find the points on the curve which satisfy this
2
condition. To do so, substitute x3 in for y in the equation for the curve:
3
x +
x2
3
3
− 9x
x2
3
=0
x6
− 3x3 = 0
27
x6
− 2x3 = 0
27
3
3 x
x
− 2 = 0.
27
x3 +
3
Therefore, the curve has horizontal tangent lines when x = 0 and when x27 − 2 = 0, meaning
√
√
2
that x = 3 54 = 3 3 2. Since these points also satisfy the condition y = x3 , we can see that
these are the points
√
√
3
3
(0, 0) and
3 2, 3 4 .
Since we’re asked for√ the √
point
which isn’t the origin, we can conclude that the point we’re
after is the point 3 3 2, 3 3 4 .
3