# A spherical balloon is being inflated. Find the rate of increase of the surface area s with...

## Question:

A spherical balloon is being inflated. Find the rate of increase of the surface area s with respect to the radius r when r = 1 ft. s = 4{eq}\pi r^2{/eq}?

## Differentiation:

The differentiation is the process of determining the rate of change of one quantity with respect to another quantity. The differentiation of a function is denoted by **'d'** operator and can be written as:

{eq}\displaystyle { v' = \frac{dv}{dt} } {/eq}

which means the rate of change of **'v'** with respect to the change in **'t'.**

The differentiation of a function at a point determines the slope of the function graph at that point. The slope gives us an idea about how fast the function graph is increasing or decreasing.

## Answer and Explanation:

The formula for the surface area of balloon is,

{eq}\displaystyle { s = 4 \pi r^2 } {/eq}

Differentiating the expression with respect to **r**:

{eq}\displaystyle { \frac{ds}{dr} = \frac{d}{dr} ( 4 \pi r^2) \\ \Rightarrow \frac{ds}{dr} = 4 \pi \frac{d}{dr} ( r^2) \\ \Rightarrow \frac{ds}{dr} = 4 \pi ( 2r) \\ \Rightarrow \frac{ds}{dr} = 8 \pi r } {/eq}

At {eq}\displaystyle { r = 1 \ \text{ft} : } {/eq}

{eq}\displaystyle { \\ \Rightarrow \frac{ds}{dr} = 8 \pi (1) = 8 \pi } {/eq}

Therefore the rate of increase of the surface area **s** with respect to the radius **r** when **r = 1 ft** is {eq}8 \pi. {/eq}

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#### Learn more about this topic:

from Math 104: Calculus

Chapter 9 / Lesson 4