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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} ...