Question

can anyone help me solve problem 2E-7 in Problem Set 4 my answer and approach is totally different form the solution.

Answer 1

Well, let's draw and write everything out. Lets start with the information they give us to draw the trapezoid. They say the base of the trapezoid is .5 meters. Since the angles are 45 degrees, that means that the two sides are the same (45/45/90 degree triangle)! So h can be put in for for the height and other side of the triangle. So here is a trapezoid with a base .5 meters with the bottom angles at 45 degrees with the heights marked.



Simple enough!

So all good there. Now, the trough is being filled at 1 cubic meter per second. That translates to the derivative of volume over time. (dv/dt = 1). The volume of a trapezoidal prism is Length * Area of the Trapezoid. They already told us the Length (4) so we have that.

Volume = 4 * (area of the trapezoid)

How they find the area of the trapezoid is pretty interesting to be honest. You'll notice that the two triangles on either side of the trapezoid have the same lengths of h. Well, put those triangles together and you get a square, and the area of the square is length*width. Therefore those triangles together is h^2. Then what they do is add the area of the rectangle in the middle by multiplying the length times width or h/2.

All together that makes \[V = 4(h^{2}+(h \div2))\ or\ V = 4h^{2}+2h\]
Now you just take the derivative of this.
\[V \prime = 8hh \prime + 2ph \prime\ or\ V \prime/(8h + 2) = h \prime\]
Now plugging in h = .5 and V prime =1, you should get h prime is equal to 1/6

Let me know if you have any questions.