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Hey there

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A hemispherical bowl of radius a has its axis vertical and is full of water. At time t=0 water starts running out of a small hole in the bottom of the bowl so that the depth of water in the bowl at t is x. The rate at which the volume of water is decreasing is proportional to x. Given that the volume of water in the bowl when the depth is x is [tex]\pi(ax^2-\frac{1}{3}x^3)[/tex] show that there is a positive constant k such that

[tex]\pi(2ax-x^2)\dfrac{dx}{dt}=-kx[/tex]

My method

[tex]-\dfrac{dV}{dt}\propto x[/tex]

[tex]\dfrac{dV}{dt}=-kx[/tex]

[tex]\dfrac{dV}{dt}=\dfrac{dV}{dx} \dfrac{dx}{dt}[/tex]

[tex]V=\pi(ax^2-\frac{1}{3}x^3)[/tex]

[tex]\dfrac{dV}{dx}=\pi(2ax-x^2)[/tex]

[tex]\pi(2ax-x^2)\dfrac{dx}{dt}=-kx[/tex]

(ii) Given that the bowl is empty after a time T, show that

[tex]k=\dfrac{3 \pi a^2}{2T}[/tex]

My method (I'm not sure how to answer this part).

[tex]\pi(2ax-x^2)\dfrac{dx}{dt}=-kx[/tex]

[tex]\displaystyle \int (2\pi a-\pi x )dx=\displaystyle \int-k dt[/tex]

[tex]2\pi ax-\frac{\pi x^2}{2}=-kt + C[/tex]

x=0 when t=T

I'm not sure how to go about showing [tex]k=\dfrac{3 \pi a^2}{2T}[/tex]with the information in the question.

Can you help me?

Thanks.

[tex]\pi(2ax-x^2)\dfrac{dx}{dt}=-kx[/tex]

My method

[tex]-\dfrac{dV}{dt}\propto x[/tex]

[tex]\dfrac{dV}{dt}=-kx[/tex]

[tex]\dfrac{dV}{dt}=\dfrac{dV}{dx} \dfrac{dx}{dt}[/tex]

[tex]V=\pi(ax^2-\frac{1}{3}x^3)[/tex]

[tex]\dfrac{dV}{dx}=\pi(2ax-x^2)[/tex]

[tex]\pi(2ax-x^2)\dfrac{dx}{dt}=-kx[/tex]

(ii) Given that the bowl is empty after a time T, show that

[tex]k=\dfrac{3 \pi a^2}{2T}[/tex]

My method (I'm not sure how to answer this part).

[tex]\pi(2ax-x^2)\dfrac{dx}{dt}=-kx[/tex]

[tex]\displaystyle \int (2\pi a-\pi x )dx=\displaystyle \int-k dt[/tex]

[tex]2\pi ax-\frac{\pi x^2}{2}=-kt + C[/tex]

x=0 when t=T

I'm not sure how to go about showing [tex]k=\dfrac{3 \pi a^2}{2T}[/tex]with the information in the question.

Can you help me?

Thanks.

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