triangular prism trough with isoceles trapezoid volume

That's dV/dt (the change in volume over the change in time). We know that the change in volume with respect to time is 2 cubic feet per minute. To do this derivation, we have to use the chain rule on the right hand side: Take the derivative of the equation with respect to time.

And because the volume of water ( V) is equal to this cross-sectional area times the length of the trough, then we have an equation relating the volume of water to the depth ( h) of water:Ģ. Since the area of the isosceles triangle is xh, this equals ( h/4) h = h 2/4.

So if we know h, we know x (and vice versa). The ratios of corresponding sides of similar triangles are equal. We can use the principle of similar triangles to relate x to h though: The area of the isoceles triangle filled with water is xh. The cross section is an isosceles triangle, of course, whose shape is defined by the relative sizes of its sides (these are given). The volume of the water in the trough equals the length of the trough times the cross-sectional area of the trough up to the depth it is filled with water. The final step is to substitute in the values you are given for the depth and the rate of volume change and you will get the rate of depth change, that's the answer to the problem.The second step is to take the derivative of both sides of the equation with respect to time.The first step is to find an equation that relates water depth to volume.This problem can be solved in three steps: You have a rate of change of volume and want to know the corresponding rate of change of depth at a particular depth. If water flows in at the rate of 2ft^3/min, how fast is the surface rising when the water is 1 ft deep ?

Related rates (a water trough) - Math CentralĪ rectangular trough is 3ft long, 2ft across the top and 4 ft deep.

Triangular prism trough with isoceles trapezoid volume