Related Rates Cone Question: A tank in the shape of a cone is being filled with water at a rate of 10m^3/s....

[matheyyy]

Hi all,

I've been struggling with this question here. I have attached my workings - are they correct? (I'm not 100% sure about the statement where I assumed that radius is always a fixed ratio with height - see attached).



QUESTION:
A tank in the shape of a cone is being filled with water at a rate of 10m^3/s. The radius of the base of the tank is 25m and the height of the tank is 15m. At what rate is the depth of the water in the tank changing when the radius of the top of the water is 10m?


Thanks everyone,
Matheyyy

 

[stapel]

I've been struggling with this question here. I have attached my workings - are they correct? (I'm not 100% sure about the statement where I assumed that radius is always a fixed ratio with height - see attached).

QUESTION:
A tank in the shape of a cone is being filled with water at a rate of 10m^3/s. The radius of the base of the tank is 25m and the height of the tank is 15m. At what rate is the depth of the water in the tank changing when the radius of the top of the water is 10m?

I will type out your work, so people can view it easily:

[imath]r=25[/imath]

[imath]h=15[/imath]

[imath]\dfrac{dV}{dt}=10[/imath]

[imath]\dfrac{dh}{dt}=?[/imath]

[imath]r=10[/imath]

[imath]V=\pi r^2 \dfrac{h}{3}[/imath]

[imath]r=\dfrac{5}{3}h[/imath]

[imath]10=\dfrac{5}{3}h[/imath]

[imath]h=6[/imath]

[imath]V=\pi \left(\dfrac{5}{3}h\right)^2 \left(\dfrac{1}{3}h\right)[/imath]

[imath]\begin{aligned} \dfrac{dV}{dt} &= \pi \left( \dfrac{5}{3} \right)(2)(h)\dfrac{dh}{dt} \cdot \dfrac{1}{3}\dfrac{dh}{dt} \\[1em] &= \pi \left(\dfrac{10}{3}\right)(h)\dfrac{dh}{dt}\cdot\dfrac{1}{3}\dfrac{dh}{dt} \\[1em] &= \pi \left(\dfrac{10}{3}\right)h \dfrac{dh}{dt} \left(\dfrac{1}{3}\right)\dfrac{dh}{dt} \\[1em] &= \dfrac{10\pi}{9}h\cdot 2\dfrac{dh}{dt} \\[1em] &= \dfrac{20\pi}{9}h\cdot \dfrac{dh}{dt} \end{aligned}[/imath]

[imath]\begin{aligned} 10 &= \dfrac{20\pi}{9} (6) \dfrac{dh}{dt} \\[1em] &= \dfrac{120\pi}{9} \dfrac{dh}{dt}\end{aligned}[/imath]

[imath]\dfrac{90}{120\pi} = \dfrac{dh}{dt}[/imath]

[imath]\dfrac{dh}{dt} = \dfrac{3}{4\pi} \textrm{m/s}[/imath]

 

[stapel]

QUESTION:
A tank in the shape of a cone is being filled with water at a rate of 10m^3/s. The radius of the base of the tank is 25m and the height of the tank is 15m. At what rate is the depth of the water in the tank changing when the radius of the top of the water is 10m?

[imath]r=25[/imath]

[imath]h=15[/imath]

[imath]\dfrac{dV}{dt}=10[/imath]

[imath]\dfrac{dh}{dt}=?[/imath]

[imath]r=10[/imath]

[imath]V=\pi r^2 \dfrac{h}{3}[/imath]

[imath]r=\dfrac{5}{3}h[/imath]

[imath]10=\dfrac{5}{3}h[/imath]

[imath]h=6[/imath]

[imath]V=\pi \left(\dfrac{5}{3}h\right)^2 \left(\dfrac{1}{3}h\right)[/imath]

At this point, you have two instances of the variable [imath]h[/imath]. Either multiply them together (after squaring), or else apply the Product Rule. I would multiply them together to get:

[imath]\qquad V = \dfrac{25\pi}{27}\, h^3[/imath]

I'm fairly certain that this will lead to a different result.

Eliz.

 

[Steven G]

You failed to square the 5/3 !!

There is something called the product rule! Ex: The derivative wrt x of x*x is (1*x)+ (x*1) = x+x = 2x. The derivative of a product is not the product of the derivative.

 

[matheyyy]

You failed to square the 5/3 !!

There is something called the product rule! Ex: The derivative wrt x of x*x is (1*x)+ (x*1) = x+x = 2x. The derivative of a product is not the product of the derivative.

Thank you for your help!!!

 

[matheyyy]

At this point, you have two instances of the variable [imath]h[/imath]. Either multiply them together (after squaring), or else apply the Product Rule. I would multiply them together to get:

[imath]\qquad V = \dfrac{25\pi}{27}\, h^3[/imath]

I'm fairly certain that this will lead to a different result.

Eliz.

Thank you so much for taking the time out of your day to write all of that - much appreciated!