# 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.

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} 