Implicit Differentiation Application: Sphere to Surface Area

Calculus Tutorial

Implicit Differentiation Application: Sphere to Surface Area


Implicit differentiation is one of the most commonly used techniques in calculus, especially in word problems. It takes advantage of the chain rule that states:

df/dx = df/dy * dy/dx

Or the fact that the derivative of one side is the derivative of the other.

Simple example:
derivative of x + y = 3 is:
(x + y)’ = (3)’
x’ + y’ = (3)’
1 + y’ = 0
y’ = -1
You wouldn’t use implicit differentiation for this question, but it comes in useful for many word problems:

Sample Problem

Remember that the formula for the surface area and volume of a sphere:

If a snowball rolling down a hill is increasing in volume at a constant rate of 5 cm^3/sec. what is the rate of change of the surface area when the radius is 10 cm.


Step 1: we differentiate the surface area to find the rate of change:

SA = 4πr^2     find the derivative
SA’ = 4π * 2r * r’
SA’ = 8π(10) * r’
SA’ = 80π * r’

Notice that we have to use the chain rule for r because we aren’t calculating rate of change as a function of radius(r), but according to time(t). SA’ = dSA/dt and r’ = dr/dt

Step 2: We don’t know what r’ is, but we do have V’. So using implicit differentiation, we can find some values. Let’s try differentiating the volume

V = 4π/3 * r^3     find the derivative
V’ = 4π/3 * 3r^2 * r’
V’ = 4π(10)^2 * r’
V’ = 400π * r’
We now have a term for r’, with a value that we know because V’ = 5.
5 = 400π * r’     solve
5/(400π) = r’
r’ = 1/(80π)

Step 3: Plug in the value

SA’ = 80π * r’
SA’ = 80π * (1/(80π))
SA’ = 1 cm^2/sec

Therefore, the surface area grows at a rate of 1 cm^2/sec when the radius is 10 cm.

BONUS: Let’s find the general formula for questions of this type:

SA’ = 4π * 2r * r’     (see Step 1)
V’ = 4π * r^2 * r’     (see Step 2)
r’ = V’/(4πr^2)     (see Step 2; also note that 4πr^2 is equal to surface area)
put together:
SA’ = 4π * 2r * V’/(4πr^2)     or simply
SA’ = 2V’/r

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