Respuesta :
Answer:
The area of the oil spill is increasing at a rate 180 π m²/s, when the radius is 30 m.
Step-by-step explanation:
Derivative Rule:
- [tex]\frac{d}{dx}x^n=nx^{n-1}[/tex]
- [tex]\frac{d}{dx}(y^n)=ny^{n-1}\frac{dy}{dx}[/tex]
Given that
The radius of the oil spill increases at a rate 1.5 m/s.
i.e [tex]\frac{dr}{dt}= 1.5\ m/s[/tex]
We need to find the rate of area increase i.e [tex]\frac{dA}{dt}[/tex] .
We know,
The area of the oil spills A = [tex]\pi r^2[/tex] ( since it spreads in circular pattern)
[tex]\therefore A=\pi r^2[/tex]
Differentiating with respect to t
[tex]\frac{dA}{dt}=\pi. 2r\frac{dr}{dt}[/tex]
Plug the value of [tex]\frac{dr}{dt} =1.5 \ m/s[/tex]
[tex]\frac{dA}{dt}=\pi. 2r(3\ m/s)[/tex]
[tex]\Rightarrow \frac{dA}{dt}=6\pi r \ m/s[/tex]
Plug r= 30 m
[tex]\Rightarrow \frac{dA}{dt}=6\pi (30\ m) \ m/s[/tex]
=180 π m²/s
The area of the oil spill is increasing at a rate 180 π m²/s, when the radius is 30 m.
