Hardest calc problem I ever saw, can you help?

[Queenisabella87]

7) A trough with a trapezoidal ends is 20 feet long, 2 feet high, 3 feet wide at the top and 2 feet wide at the bottom.

If water is being pumped in a rate of 2ft3 / min. How was fast is the depth increasing when the water is 1 foot deep.

(express your answer in inches / minute )




I'm struggling with this one terribly, don't know where to begin. :!:

Not just looking for the answer but explanation if possible.

 

[galactus]

This is a case of more simialr triangles.

\(\displaystyle \frac{s}{h}=\frac{1}{2}\)

\(\displaystyle s=\frac{h}{2}\)

\(\displaystyle \frac{dV}{dt}=2\)

Width of water a depth h is \(\displaystyle w=2+2s=2+2(\frac{h}{2})=2+h\)

\(\displaystyle V=\frac{20}{2}(2+2+h)h=20h(h+1)\)

\(\displaystyle \frac{dV}{dt}=(40h+20)\frac{dh}{dt}\)

You have dV/dt and h, solve ofr dh/dt.

 

[Queenisabella87]

thank yoU!!! =)

 

[Queenisabella87]

hey im sorry , what is h = ? thats all I need

 

[Deleted member 4993]

hey im sorry , what is h = ? thats all I need

Ms. Queen,

Either you are extremely tired or you are not ready for this class.

Are these parts of take-home exam?

You are posting different types of problems that require thought and understanding - and as day passes by I see your questions are becoming less thought-ful. In the problem above you need to find "rate" of change of 'h' - not 'h'. In your problem statement the 'h' is already given.