A trough is triangular in cross section, an isosceles triangle with sides of 12 inches, and a top of 10 inches. The trough is 40 inches long. How fast is the depth changing if you are pumping one cubic foot per minute into the trough?
I am having a hard time getting it in terms of h.
Make a sketch of the cross-section of the isosceles triangle. Draw it some water level, letting the height be h and letting the width of the water level be x
by similar triangles
x/h = 10/12
12x = 10h
x = 5h/6
volume of water = 40(area of triangle)
= 40(1/2)(x)h
= 20(5h/6)(h) = 50/3 h^3
V = (50/3) h^3
dV/dt = 50 h^2 dh/dt
have to watch our units, 1 cubic foot = 12^3 or 1728 cubic inches
1728 = 50 h^2 dh/dt
dh/dt = 1728/(50h^2) = 864/25 inches/minute
wow I was headed that way didn't watch units Thank You very much
To find how fast the depth of the trough is changing, we need to use the concept of related rates. Let's denote the depth of the trough as "h" (in inches) and the volume of water in the trough as "V" (in cubic inches).
We are given that the trough is triangular in cross-section, with an isosceles triangle of sides 12 inches. This means the base of the triangle is also 12 inches. Moreover, the top of the triangle (which is parallel to the base) has a length of 10 inches.
The volume of the trough is given by the formula V = (1/2) * base * height * length, where the height represents the depth of the trough. Since the base and length are constant at 12 inches and 40 inches, respectively, we can express the volume in terms of the height (h) as:
V = (1/2) * 12 * h * 40
V = 240h
We are told that water is pumped into the trough at a rate of 1 cubic foot per minute, which is equivalent to 12^3 cubic inches per minute. So the rate of change of the volume with respect to time (dV/dt) is 12^3 cubic inches per minute.
The problem asks for the rate of change of the depth, dh/dt. To find this, we need to differentiate the volume equation with respect to time using implicit differentiation:
dV/dt = d(240h)/dt
12^3 = 240 * dh/dt
Now, we can solve for the rate of change of the depth:
dh/dt = (12^3) / 240
Simplifying this expression, we have:
dh/dt = 6 in/min
Therefore, the depth is changing at a rate of 6 inches per minute.