question 14\nthe radius of a circular oil spill after t minutes is given by $r(t)=sqrt{25t}$. find the…

question 14\nthe radius of a circular oil spill after t minutes is given by $r(t)=sqrt{25t}$. find the instantaneous rate at which the radius is growing after 27 minutes.\ngive your answer as a decimal approximation with at least 3 decimal places.\nquestion help: video

question 14\nthe radius of a circular oil spill after t minutes is given by $r(t)=sqrt{25t}$. find the instantaneous rate at which the radius is growing after 27 minutes.\ngive your answer as a decimal approximation with at least 3 decimal places.\nquestion help: video

Answer

Explanation:

Step1: Rewrite the function

Rewrite $r(t)=\sqrt{25t}=5\sqrt{t}=5t^{\frac{1}{2}}$.

Step2: Differentiate the function

Using the power - rule $\frac{d}{dt}(t^n)=nt^{n - 1}$, we have $r^\prime(t)=5\times\frac{1}{2}t^{\frac{1}{2}-1}=\frac{5}{2}t^{-\frac{1}{2}}=\frac{5}{2\sqrt{t}}$.

Step3: Evaluate at $t = 27$

Substitute $t = 27$ into $r^\prime(t)$. So $r^\prime(27)=\frac{5}{2\sqrt{27}}=\frac{5}{2\times3\sqrt{3}}=\frac{5}{6\sqrt{3}}$. Rationalize the denominator: $r^\prime(27)=\frac{5\sqrt{3}}{18}\approx0.481$.

Answer:

$0.481$