Answer:
929.91 cm²/min
461.59 ft²/min
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{4cm}\underline{Area of a circle}\\\\$A=\pi r^2$\\\\where:\\ \phantom{ww}$\bullet$ $r$ is the radius. \\ \end{minipage}}[/tex]
Differentiate the expression for area with respect to r:
[tex]\implies \dfrac{\text{d}A}{\text{d}r}=2\pi r[/tex]
Given the radius is increasing at a rate of 4 centimeters per minute:
Use the chain rule to find an expression for the rate of change in the area with respect to time:
[tex]\implies \dfrac{\text{d}A}{\text{d}t}=\dfrac{\text{d}A}{\text{d}r} \times \dfrac{\text{d}r}{\text{d}t}[/tex]
[tex]\implies \dfrac{\text{d}A}{\text{d}t}=2 \pi r \times 4[/tex]
[tex]\implies \dfrac{\text{d}A}{\text{d}t}=8 \pi r[/tex]
When r = 37 cm:
[tex]\implies \dfrac{\text{d}A}{\text{d}t}=8\pi (37)[/tex]
[tex]\implies \dfrac{\text{d}A}{\text{d}t}=296 \pi[/tex]
[tex]\implies \dfrac{\text{d}A}{\text{d}t}=929.91\; \sf (2\;d.p.)[/tex]
Therefore, the rate of change in the area when the radius is 37 centimeters is 929.91 cm²/min.
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[tex]\boxed{\begin{minipage}{5 cm}\underline{Area of an equilateral triangle}\\\\$A=\dfrac{\sqrt{3}}{4}s^2$\\\\where:\\ \phantom{ww}$\bullet$ $s$ is the side length. \\ \end{minipage}}[/tex]
Differentiate the expression for area with respect to s:
[tex]\implies \dfrac{\text{d}A}{\text{d}s}=\dfrac{2\sqrt{3}}{4}s=\dfrac{\sqrt{3}}{2}s[/tex]
Given the length of each side is increasing at a rate of 13 feet per hour.
[tex]\implies \dfrac{\text{d}s}{\text{d}t}=13[/tex]
Use the chain rule to find an expression for the rate of change of the area with respect to time:
[tex]\implies \dfrac{\text{d}A}{\text{d}t}=\dfrac{\text{d}A}{\text{d}s} \times \dfrac{\text{d}s}{\text{d}t}[/tex]
[tex]\implies \dfrac{\text{d}A}{\text{d}t}=\dfrac{\sqrt{3}}{2}s \times 13[/tex]
[tex]\implies \dfrac{\text{d}A}{\text{d}t}=\dfrac{13\sqrt{3}}{2}s[/tex]
When s = 41 ft:
[tex]\implies \dfrac{\text{d}A}{\text{d}t}=\dfrac{13\sqrt{3}}{2}(41)[/tex]
[tex]\implies \dfrac{\text{d}A}{\text{d}t}=\dfrac{533\sqrt{3}}{2}[/tex]
[tex]\implies \dfrac{\text{d}A}{\text{d}t}=461.59 \sf \;\;(2\;d.p.)[/tex]
Therefore, the rate of change in the area when the side length is 41 feet is 461.59 ft²/min.
Differentiation rules used:
[tex]\boxed{\begin{minipage}{4.5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=nx^{n-1}$\\\end{minipage}}[/tex]