Stitz-Zeager_College_Algebra_e-book

Theorem 52 tells us that the domain of k 1 is the

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Unformatted text preview: of a sphere is a function of its radius r and is given by the formula S (r) = 4πr2 . Suppose the sphere is being inflated so that the radius of the sphere is increasing according to the formula r(t) = 3t2 , where t is measured in seconds, t ≥ 0, and r is measured in inches. Find and interpret (S ◦ r)(t). Solution. If we look at the functions S (r) and r(t) individually, we see the former gives the surface area of a sphere of a given radius while the latter gives the radius at a given time. So, given a specific time, t, we could find the radius at that time, r(t) and feed that into S (r) to find the surface area at that time. From this we see that the surface area S is ultimately a function of 2 time t and we find (S ◦ r)(t) = S (r(t)) = 4π (r(t))2 = 4π 3t2 = 36πt4 . This formula allows us to compute the surface area directly given the time without going through the ‘middle man’ r. A useful skill in Calculus is to be able to take a complicated function and break it down into a composition of easier functions w...
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