A spherical balloon is being inflated. Find the rate (in [invalid MathML mathml] / [invalid MathML mathml]) of increase of the surface area ([tex]S=4 \pi r^2[/tex]) with respect to the radius [tex]r[/tex] when [tex]r[/tex] is each of the following.
(a) [tex]1 f[/tex]
(b) [tex]3 f[/tex]
(c) [tex]8 f[/tex]
Answer (1)
Find the derivative of the surface area formula S = 4 π r 2 with respect to r , which gives d r d S = 8 π r .
Evaluate the derivative at r = 1 , r = 3 , and r = 8 .
At r = 1 , the rate of increase is 8 π .
At r = 3 , the rate of increase is 24 π , and at r = 8 , the rate of increase is 64 π . Therefore, the answers are 8 π , 24 π , 64 π .
Explanation
Problem Setup We are given the formula for the surface area of a sphere, S = 4 π r 2 , and we want to find the rate of increase of the surface area with respect to the radius, which is the derivative d r d S .
Finding the Derivative First, we need to find the derivative of S with respect to r . Using the power rule, we have d r d S = d r d ( 4 π r 2 ) = 4 π d r d ( r 2 ) = 4 π ( 2 r ) = 8 π r .
Evaluating at Given Radii Now we need to evaluate d r d S at the given values of r :
(a) When r = 1 , d r d S = 8 π ( 1 ) = 8 π .
(b) When r = 3 , d r d S = 8 π ( 3 ) = 24 π .
(c) When r = 8 , d r d S = 8 π ( 8 ) = 64 π .
Final Results So, the rates of increase of the surface area with respect to the radius at r = 1 , r = 3 , and r = 8 are 8 π , 24 π , and 64 π , respectively.
Examples
Imagine you're inflating a balloon and want to know how quickly the surface area is growing as you increase the radius. This problem helps you calculate that rate. For instance, knowing how the surface area changes with the radius is crucial in designing balloons or other inflatable structures, ensuring they can withstand the pressure and expand as expected. Understanding these rates is also fundamental in various engineering applications, such as designing spherical tanks or understanding heat transfer from spherical objects.
