A growing sand pile Sand falls from a conveyo...

Question

A growing sand pile Sand falls from a conveyor belt at the rate of 10 m³/min onto the top of a conical pile. The height of the pile is always three-eighths of the base diameter. How fast are the (a) height and (b) radius changing when the pile is 4 m high? Answer in centimeters per minute.

Accepted Answer

A spherical balloon is being inflated such that its volume increases at a rate of 50 centimeters cubed per second. If the radius of the balloon is always equal to half the diameter, how fast is the A radius, and B, surface area increasing when the radius is 10 centimeters? Now, to solve this, we will use the volume and surface area equations for a sphere. Volume of a sphere. Is given By 4/3. Pyar cuts. And the surface area of the sphere. Is given by 4 pi are squared. Now, to solve this, we need to take the derivative in terms of time on both sides of the equation. Let's first look at volume. We have DVDT equals the derivative in terms of volume on the right side, and radius. We have 4/3 pie. And then the derivative of R cubed is 3R2d. And because this is implicitly differentiated, we get a DRDT after. Now, it tells us the volume increases at a rate of 50 centimeters cubed per second. So we have 50 centimeters cubed per second, equals 4/3 pi multiplied by 3R2 DRDT. Now we're solving for DRDT. So, We have 3 multiplied by R, which in our case is 10 centimeters. Squared TRDT. You get 50 equals 4/3 pi. Multiplied By 300 centimeters squared, DR. DT Now we just need to simplify. We end up getting 50 equals. 4 pi multiplied by 100 T R D T. Which is just 50 divided by 400. I equals DRDT. We can simplify even further to get 1 divided by 8 pi. centimeters per second. That is our change in radius. Now we need to see our change in surface area. So you know the surface area equation is 4 pi squared. And then our derivative. It is D S A D T. Which will be 4 pi multiplied by 2R. DR DT. Which is 8IR TRDT. Now D S A D T equals Api multiplied by radius. 10 centimeters, and DRDT, which is the answer we got previously, 1 divided by 8 i. We can now simplify. We noticed 8 pi and 1 divided by 8 pi cancel each other out, just giving us 10. centimeters, and because this is changed in surface area, this is centimeters squared per second. We now have our two answers. One divided by 8 centimeters per second. And 10 centimeters squared per second. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

A growing sand pile Sand falls from a conveyor belt at the rate of 10 m³/min onto the top of a conical pile. The height of the pile is always three-eighths of the base diameter. How fast are the (a) height and (b) radius changing when the pile is 4 m high? Answer in centimeters per minute.

Key Concepts

Related Rates

Volume of a Cone

Geometric Relationships

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