# Related Rates-Calculus

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• Mar 23rd 2009, 12:44 PM
Jim Marnell
Related Rates-Calculus
A trough has a triangular cross section. The trough is 6ft across the top, 6ft deep, and 16ft long. Water is being pumped into the trough at the rate of 4ft^3 per minute. Find the Rate at which the height of the water is increasing at the instant that the height is 4ft.

Not even sure what to do.
• Mar 23rd 2009, 01:25 PM
earboth
Quote:

Originally Posted by Jim Marnell
A trough has a triangular cross section. The trough is 6ft across the top, 6ft deep, and 16ft long. Water is being pumped into the trough at the rate of 4ft^3 per minute. Find the Rate at which the height of the water is increasing at the instant that the height is 4ft.

Not even sure what to do.

The volume of the water is:

$V=\dfrac12 \cdot w \cdot h \cdot 16 = 8wh$

Use proportions:

$\dfrac wh=\dfrac 66~\implies~w = h$

Therefore:

$V(h)=8h^2$

Differentiate both sides wrt t:

$\dfrac{dV}{dt} = 16h \cdot \dfrac{dh}{dt}$

Plug in the values you know:

$4 \dfrac{cft}{min} = 16 \cdot 4 \cdot \dfrac{dh}{dt}$

Solve for $\dfrac{dh}{dt}$. I've got $\dfrac{dh}{dt}=\dfrac1{16} \dfrac{ft}{min}$
• Mar 23rd 2009, 02:43 PM
Jim Marnell
Thanks for the help! The picture helped understand it more too.