Related Rates Problem

[revisisland24]

Hey guys,

I'm working on a review packet for my calc final and was wondering if my answers are correct. Here's the question:

Suppose that water is being pumped into a trough. The trough is 15 feet long, and every cross section of the trough is an equilateral triangle with a vertex of the triangle pointing towards the ground. Find the rate of change of the height of the water when the height is 4 ft., if the volume is increasing at a rate of 8 cubic ft./sec.

I got that v = 1/2bh and that V' = 8. I know that r'(h) = 4 ft, but am having trouble putting the info together. Can anyone point me in the right direction?

 

[daon2]

Imagine you are facing the front of this tank so as to see the equilateral triangle. Put the origin at the bottom vertex. Then the sides of the triangle are lines coming out of the origin and have slope -sqrt(3) and sqrt(3) respectively (equilaterals have angles of 60 degrees, and with a little analysis you can get here). And yes, as you have found, V=1/2*(base)*(height)*(length).

The equation of the positive-sloped line coming out of your origin is h=sqrt(3)*x. When the water is at height h, half its base will be x=h/sqrt(3). By symmetry, the entire base is 2h/sqrt(3).

So V = h^2/sqrt(3)*15.

Now differentiate with respect to time.

 

[MarkFL]

I would let the base of the inverted equilateral triangular cross-section by \(\displaystyle b\), the depth of the water be \(\displaystyle h\) and the length of the trough be \(\displaystyle \ell\) (this is a constant).

Looking at the cross-section, we may state:

\(\displaystyle \tan(60^{\circ})=\dfrac{h}{\dfrac{b}{2}}\)

\(\displaystyle \sqrt{3}=\dfrac{2h}{b}\)

\(\displaystyle b=\dfrac{2}{\sqrt{3}}h=\dfrac{2\sqrt{3}}{3}h\)

Now, the volume of the water in the trough is:

\(\displaystyle V=\dfrac{1}{2}bh\ell=\dfrac{\sqrt{3}}{3}\ell h^2\)

Now, implicitly differentiate this with respect to \(\displaystyle t\), solve for \(\displaystyle \dfrac{dh}{dt}\), and plug in the give values for \(\displaystyle \dfrac{dV}{dt}\) and \(\displaystyle \ell\).

 

[revisisland24]

Imagine you are facing the front of this tank so as to see the equilateral triangle. Put the origin at the bottom vertex. Then the sides of the triangle are lines coming out of the origin and have slope -sqrt(3) and sqrt(3) respectively (equilaterals have angles of 60 degrees, and with a little analysis you can get here). And yes, as you have found, V=1/2*(base)*(height)*(length).

The equation of the positive-sloped line coming out of your origin is h=sqrt(3)*x. When the water is at height h, half its base will be x=h/sqrt(3). By symmetry, the entire base is 2h/sqrt(3).

So V = h^2/sqrt(3)*15.

Now differentiate with respect to time.

Then, V'(t) = 30h/sqrt(3) * dh/dt then i could write 8= 30(4)/sqrt(3)dh/dt and dh/dt = 1/15 * sqrt (3). Is that right?

 

[revisisland24]

I would let the base of the inverted equilateral triangular cross-section by \(\displaystyle b\), the depth of the water be \(\displaystyle h\) and the length of the trough be \(\displaystyle \ell\) (this is a constant).

Looking at the cross-section, we may state:

\(\displaystyle \tan(60^{\circ})=\dfrac{h}{\dfrac{b}{2}}\)

\(\displaystyle \sqrt{3}=\dfrac{2h}{b}\)

\(\displaystyle b=\dfrac{2}{\sqrt{3}}h=\dfrac{2\sqrt{3}}{3}h\)

Now, the volume of the water in the trough is:

\(\displaystyle V=\dfrac{1}{2}bh\ell=\dfrac{\sqrt{3}}{3}\ell h^2\)

Now, implicitly differentiate this with respect to \(\displaystyle t\), solve for \(\displaystyle \dfrac{dh}{dt}\), and plug in the give values for \(\displaystyle \dfrac{dV}{dt}\) and \(\displaystyle \ell\).

Then, V'(t) = 30h/sqrt(3) * dh/dt then i could write 8= 30(4)/sqrt(3)dh/dt and dh/dt = 1/15 * sqrt (3). Is that right?

 

[MarkFL]

Yes, that's correct, in ft/s.

 

[revisisland24]

Yes, that's correct, in ft/s.

thanks!