## Calculus Tutorial

*Implicit Differentiation Application: Sphere to Surface Area*

#### Intro

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.

#### Solution

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**

# About The Author

University Student; Biomedical Sciences |

Outstanding Math, Chemistry, and Physics student (high 90's). Solid understanding of the foundations of concepts taught in the Alberta high school curriculum, along with its applications. General experience tutoring peers. Experienced in leading and coordinating high school students. Mentor of h... |

## Leave a Comment