13. -/2 points details my notes scalc9 3.1.523.xp.mi. find the absolute minimum and absolute maximum values…

Explanation:

Step1: Find the derivative of (y = t\sqrt{9 - t^{2}})

Use the product - rule ((uv)^\prime=u^\prime v + uv^\prime), where (u = t) and (v=\sqrt{9 - t^{2}}=(9 - t^{2})^{\frac{1}{2}}). (u^\prime = 1), (v^\prime=\frac{1}{2}(9 - t^{2})^{-\frac{1}{2}}\times(- 2t)=\frac{-t}{\sqrt{9 - t^{2}}}). So (y^\prime=\sqrt{9 - t^{2}}+t\times\frac{-t}{\sqrt{9 - t^{2}}}=\frac{9 - t^{2}-t^{2}}{\sqrt{9 - t^{2}}}=\frac{9 - 2t^{2}}{\sqrt{9 - t^{2}}}).

Step2: Find the critical points

Set (y^\prime = 0), then (\frac{9 - 2t^{2}}{\sqrt{9 - t^{2}}}=0). The numerator must be zero, so (9 - 2t^{2}=0), which gives (t^{2}=\frac{9}{2}), (t=\pm\frac{3}{\sqrt{2}}). But we are in the interval ([-1,3]), so we consider (t = \frac{3}{\sqrt{2}}) (since (t=-\frac{3}{\sqrt{2}}\notin[-1,3])). Also, the derivative is undefined when (9 - t^{2}=0), i.e., (t = 3) or (t=- 3), but (t=-3\notin[-1,3]).

Step3: Evaluate the function at critical points and endpoints

Evaluate (y(t)=t\sqrt{9 - t^{2}}) at (t=-1), (t=\frac{3}{\sqrt{2}}), and (t = 3). When (t=-1), (y(-1)=-1\times\sqrt{9 - 1}=- \sqrt{8}=-2\sqrt{2}). When (t=\frac{3}{\sqrt{2}}), (y(\frac{3}{\sqrt{2}})=\frac{3}{\sqrt{2}}\times\sqrt{9-\frac{9}{2}}=\frac{3}{\sqrt{2}}\times\frac{3}{\sqrt{2}}=\frac{9}{2}). When (t = 3), (y(3)=3\times\sqrt{9 - 9}=0).

Answer:

absolute minimum: (-2\sqrt{2}) absolute maximum: (\frac{9}{2})