Related rates grain hopper problem

[decamij]

Grain is emptying from the bottom of a funnel-shaped hopper at a rate of 1.2m3/min. If the diameter of the top of the hopper is 5m and the sides make an angle of 30o with the vertical, determine the rate at which the level of grain in the hopper is changing, when it is half the height of the hopper.
P.S. (I am supposed to use implicit differentiation to find rates of change)

The answer is -0.24m/min.

I can't get it!

 

[BobG]

You can use explicit differentiation.

The angle is 30 degrees. The tangent of 30 degrees equals the radius divided by the heighth. In other words:

$$h=\frac{r}{tan 30}$$

The angle stays constant, so tan 30 is also a constant.

As the grain empties, it's cross section is always a similar triangle (similar to the shape of the hopper). If the heighth of the hopper is one-half, so are all the other sides.

Edit: I thought you were asking if you had to use implicit differentiation.

The 1/3 pi is just a constant. You apply the product rule to differentiate hr^2 (keeping in mind you have to apply the chain rule, as well).

To get an answer, you still have to eventually find h from r.

 

[M98Ranger]

Don't worry, finding the rate of change in related rates problems can be tricky at first. Let's break down the problem and go through the steps to solve it using implicit differentiation.

First, we need to identify what we are given and what we are looking for. We are given the rate at which grain is emptying from the bottom of the hopper (1.2m3/min), the diameter of the top of the hopper (5m), and the angle of the hopper's sides (30o). We are looking for the rate at which the level of grain in the hopper is changing when it is half the height of the hopper.

Next, we need to draw a diagram to visualize the problem. From the given information, we can draw a right triangle with the height of the hopper (h) as the hypotenuse, the base of the triangle (b) as half the diameter (2.5m), and the angle of 30o. We can also label the rate of change we are looking for as dh/dt.

Now, we can set up an equation using the volume of a cone formula, V = (1/3)πr2h, where r is the radius of the base of the hopper. Since we know the diameter of the top of the hopper is 5m, the radius is 2.5m. We can also rewrite the equation in terms of h, V = (1/3)π(2.5)2h.

To use implicit differentiation, we need to take the derivative of both sides of the equation with respect to time (t). This means we need to use the chain rule and the product rule. The derivative of V with respect to t is dV/dt, and the derivative of h with respect to t is dh/dt. The derivative of (1/3)π(2.5)2h with respect to t is (1/3)π(2.5)2(dh/dt).

Now, we can plug in the given rate of change for the volume (1.2m3/min) and the value for the radius (2.5m) into the equation. We also know that when the level of grain is half the height of the hopper, the value of h is 2.5m (since the height of the hopper is 5m). This gives us the