Question A mathematician stands on a beach with his dog at point A. He throws a tennis ball so that it hits the water at point B. The dog, wanting to get to the tennis ball as quickly as possible, runs along the straight beach line to point D and then swims from point D to point B to retrieve his ball. Assume C is the point on the edge of the beach closest to the tennis ball (see figure). Complete parts (a) through (d) below. a. Assume the dog runs at speed r and swims at speed s, where r>s and both are measured in meters/second. Also assume the lengths of BC, CD, and AC are x, y, and z, respectively. Find a function T(y) representing the total time it takes for the dog to get to the ball. T(y) =O 3D

Given that a mathematician stands on a beach with his dog at point A.He throws ball and it hits the water at point \mathrm{B}. The dog runs along the straight line to \mathrm{D} and then swims to point \mathrm{B} to get the ball.To Find: If the dog runs at speed r and swims at speed s, such that r>s, determine the total time it takes for the dog to get the ball represented by function \mathrm{T}(\mathrm{y}).Begin with finding DB,From the figure, we get,\mathrm{AC}=\mathrm{z}, \mathrm{BC}=\mathrm{x}, \mathrm{DC}=\mathrm{y}Thus, \mathrm{AD}=\mathrm{z}-\mathrm{y}Using the Pythagoras theorem in DCB,\begin{aligned}\mathrm{x}^{2}+\mathrm{y}^{2} & =\mathrm{DB}^{2} \\\therefore \mathrm{DB} & =\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}\end{aligned}Further, the dog travelled distance \mathrm{AD} at speed \mathrm{r} \mathrm{m} / \mathrm{s} and distance \mathrm{DB} at a speed \mathrm{s} \mathrm{m} / \mathrm{s}We calculate the total time using the formula for speed which is\begin{aligned}\text { speed } & =\frac{\text { distance }}{\text { time }} \\\mathrm{T}(\mathrm{y}) & =\frac{\mathrm{AD}}{\mathrm{r}}+\frac{\mathrm{DB}}{\mathrm{s}} \\& =\frac{\mathrm{z}-\mathrm{y}}{\mathrm{r}}+\frac{\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}}{\mathrm{~s}}\end{aligned}Therefore,T(y)=\frac{z-y}{r}+\frac{\sqrt{x^{2}+y^{2}}}{s} ...