Today, we will be discussing some more examples involving implicit differentiation.
Problem 0.1. A cylinder is getting taller and thinner, but the surface area is
dr
held constant at 50cm2 . Find dh
when the radius is equal to 2cm.
The surface area of a cylinder is 2πr2 + 2πrh. So we take the derivative with
respect to h to get
dr
dr
+ 2πh
+ 2πr = 0.
4πr
dh
dh
Therefore, we get
dr
−r
=
.
dh
2r + h
But when r = 2, we have
8π + 4πh = 50
so
h=
50 − 8π
.
4π
We now plug in r and h to get
−2
dr
.
=
dh
4 + 50−8π
4π
Again, it’s important to take the derivative before trying to plug in any
numbers. Otherwise, you will get the wrong answer.
Problem 0.2. Suppose we are on the curve y 2 = x3 + 5x − 6.
• Are there any places where the curve has a vertical tangent?
• Find the second derivative of y with respect to x at an arbitrary place on
the curve.
2y
dy
= 3x2 + 5
dx
3x2 + 5
dy
=
dx
2y
Vertical tangent lines can only occur if y = 0. Clearly (1, 0) is a point on the
curve for which y = 0. Let’s check if there are any others: factoring gives that
x3 + 6x − 6 = (x − 1)(x2 + x + 6), and x2 + x + 6 is irreducible because the
discriminant b2 − 4ac is negative. Therefore, x = 1, y = 0 is the only possibility.
To verify that there is actually a vertical tangent line here, we need to
2y
compute dx
dy , which comes out to be 3x2 +5 , which is zero at (1, 0). So there is
indeed a vertical tangent line at (1, 0).
1
Now, let’s work on finding the second derivative. We have
2y
dy
= 3x2 + 5.
dx
Let’s take another derivative:
dy
2
dx
So we solve for
2
+ 2y
d2 y
= 6x
dx2
d2 y
dx2 :
2
d y
=
dx2
and plug in our expression for
y:
2
dy
dx
d y
=
dx2
6x − 2
2y
dy
dx
2
to get this whole thing in terms of just x and
6x − 2
3x2 +5
2y
2y
2
.
Problem 0.3. We will do another geometric example now: An ellipse can be
described as the set of points P in the plane such that, for two special points
called the foci, the sum of the distance from P to the foci is a constant. Consider
dy
the ellipse with foci (−3, 0) and (3, 0) with distance-sum equal to 10. Find dx
at the points where x = 3.
We have:
p
p
(x − 3)2 + y 2 + (x + 3)2 + y 2 = 10.
Upon differentiating, we get:
1
dy
dy
1
+p
.
2(x − 3) + 2y
2(x + 3) + 2y
0= p
dx
dx
2 (x − 3)2 + y 2
(x + 3)2 + y 2
p
At x = 3, we have ±y + 36 + y 2 = 10. A little algebra gives that y = ± 64
20 .
dy
.
We can plug this into the expression for dx
2