Related Rates Calculus Cone problem

In summary, the man is drinking soda at a rate of 20 cm^3/s. At 10 cm, the level of the soda is dropping at a rate of 20 cm^3/s.
  • #1
Batmaniac
24
0
I'm stuck on this question:

"A man is sipping soda through a straw from a conical cup, 15 cm deep and 8 cm in diameter at the top. When the soda is 10 cm deep, he is drinking at the rate of 20 cm^3/s. How fast is the level of the soda dropping at that time?"

So you are given height = 15 cm, radius = 4 cm, and the derivative of the volume at height = 10 cm is 20 cm^3/s. It would appear that the question is asking for the derivative of height at 10 cm.

So volume of a cone is 1/3*pi*r^2*h, meaning, the derivative of that is:

dV/dt = (2*pi*r*dr/dt*h)/3 + (dh/dt*pi*r^2)/3

The only problem is that to find dh/dt, as the question is asking, we need to know dr/dt, and I can't think of anything I could do to find it. So perhaps I went about this question the wrong way or there is something I'm not seeing. Any help or guidance would be greatly appreciated, thanks.
 
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  • #2
Can you find the height in terms of the radius and angle between the base and the hypotenuse formed by the slanted side of the cone?
 
  • #3
Since I'm fairly certain the angle between the base and the hypotenuse stays constant as 'h' decreases (and thus 'r' decreases), I can use that to find the radius at h = 10, which now that I think about it, was another unknown in the initial question that I did not have, but that still doesn't help me find dr/dt.
 
  • #4
I think I know what to do now.

When the height is 15, the radius is 5, since the angle between slant and base is constant, we have similar triangles, so the sides are proportional. So 1/3h = r. So we can sub in h into the original volume equation for r and then differentiate the equation so that our differentiated equation only has the one unknown, dh/dt and we can easily solve for it.
 
  • #5
Yeah I got the answer.

Math rules.
 
  • #6
Batmaniac said:
Yeah I got the answer.

Math rules.
Good, I agree :approve:
 
  • #7
It should definitely be noted that 90% of related rates questions use maybe 3 different easy geometric identities... if you just remember them, every question will end up asking the exact same thing, just with different numbers
 

FAQ: Related Rates Calculus Cone problem

1. What is a related rates problem in calculus?

A related rates problem in calculus is a type of problem that involves finding the rate of change of one quantity with respect to another, where the two quantities are related by an equation. These types of problems often involve geometric shapes, such as cones, and require the use of derivatives to solve.

2. How do you set up a related rates problem involving a cone?

To set up a related rates problem involving a cone, you will need to use the formula for the volume of a cone: V = (1/3)πr²h, where r is the radius of the base and h is the height. You will also need to identify which quantities are changing and which are constant, and then use the chain rule to find the relationship between the rates of change.

3. What is the general process for solving a related rates problem involving a cone?

The general process for solving a related rates problem involving a cone is as follows:

  1. Draw a diagram and label all given quantities and the rates of change.
  2. Write down the formula for the quantity you are trying to find, in terms of the given quantities.
  3. Differentiate the formula with respect to time, using the chain rule.
  4. Substitute in the given values and solve for the unknown rate of change.

4. What are some common mistakes to avoid when solving a related rates problem involving a cone?

Some common mistakes to avoid when solving a related rates problem involving a cone include mixing up the variables, not differentiating correctly, and not setting up the equation correctly. It is important to carefully label all given quantities and rates of change, and to double check all calculations.

5. Can you provide an example of a related rates problem involving a cone?

Sure, here is an example of a related rates problem involving a cone:

A cone-shaped paper cup is being filled with water at a rate of 15 cm³/min. If the height of the cup is 12 cm and the radius of the base is 4 cm, at what rate is the water level rising when the water is 6 cm deep?

To solve this problem, we will use the formula for the volume of a cone and the chain rule to find the relationship between the rate of change of the height and the rate of change of the radius. After substituting in the given values, we can solve for the unknown rate of change, which in this case is the rate at which the water level is rising, and it turns out to be 5 cm/min.

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