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Question 9

This version was saved 14 years, 12 months ago View current version     Page history
Saved by Rence
on April 22, 2009 at 11:18:35 pm
 

A Calculator MAY NOT be used to help answer this question.

 

Water is draining at the rate of 48π ft3/minute from the vertex at the bottom of a conical tank whose diameter at its base is 40 feet and whose height is 60 feet.

 

(a) Find an expression for the volume of water in the tank in terms of its radius at the surface of the water.

 

(b) At what rate is the radius of the water in the tank shrinking when the radius is 16 feet?

 

(c) How fast is the height of the water in the tank dropping at the instant that the radius is 16 feet?

 

Solution

 

Before you start onto the actual questions, I'd suggest drawing a diagram to get a better understanding of what is occurring. This is especially helpful for visual learners.

 

 

Now let's begin...

 

a) Since it is asking for the volume in terms of the radius, that would mean that the only variable that will be in our equation will be r. The trick to this though is that its asking in terms of the radius with respect to the water's surface. Due to this, we will need to refer to the use of similar triangles, in order to find missing varible because as shown in the diagram, we do not know the exact values for h and r. Let us write down what we have so far.

 

 

As you can see, I have solved for h using similar triangles because as stated earlier, the question is specifically asking for the volume to be in terms of the radius, r. If it had been asked to be in terms of the height, then I would have solved for r instead, leaving just the variable, h, in the equation.

 

Now knowing our value for the height, we can now substitute it into the equation for the volume of a cone. Finishing off the question, we should now be left with...

 

And there you have it, the expression for the volume of the water in the tank in terms of its radius at the surface of the water. Magnifique!

 


 

(b) Now we get to the actual related rates part of the question. First off, we should list what we are given/know and what we are looking for...

 

As you can see, we will be needing the equation for volume that we found in part a since we will later be implicitly differentiating it to get and solve for the rate of change of the radius. We were also given the rate that the volume is changing with respect to time (in minutes). The rate that is unknown to us, or in other words what we are looking for, is what we are trying to solve for in this question. The Leibniz notation standing for the change in the radius with respect to time, which is exactly what the question is asking for.

 

Now, the first step we should take is to implicitly differentiate our equation for volume. We will have to write in terms of Leibniz notation since when dealing with related rates problems, we want to be able to see the rate in which we are solving for. After implicitly differentiating we should end up with...

 

 

Iffy on implicit differentiation (I admit, I was too before I started doing this. >__>)? Well, here's a helpful video. This guy is fantastic, seriously. Check out his other videos as well if you'd like since he covers a few other calculus related subjects.

 

YouTube plugin error

 

Moving on, as you can see, we can now substitute in the values we know to solve for dr/dt. From here on out its just basic algebra.

 

Other than the large fraction to reduce, I'm sure you're also wondering how you could possibly square a two digit number without a calculator as well. Of course there's the common method of just writing it out yourself but you can use this trick which I have also mentioned on the class blog in one of my previous scribe posts. For your convenience I shall repost that trick.

 

Squaring Two Digit Numbers Without a Calculator:

 

  1. Take the nearest multiple of 10 of the number you are trying to square.
  2. After finding the nearest multiple, take the intial value you were trying to square and subtract the nearest multiple of 10 you found from that value.
  3. Then take the difference from the last step and add it to the value you were trying to square.
  4. Now take the value of your nearest multiple of 10 and the sum you got from the last step and multiply them together.
  5. Square the value you found from Step 2 and square it. It shouldn't be hard considering the fact that its only one-digit.
  6. Add the values from Steps 4 and 6 together and you are done.

 

Here's an example using the value from our question here:

 

 

Presto! Let's move on shall we?


 

 

(c) This question is pretty much the same as part b. The only difference being is that we are looking for the change in the height given the same radius from part b. The first step we should take is to write out a volume equation with respect to the height. We will need to follow the same steps we took in part a, but this time apply the focus to the height, h.

 

Continuing the same process as part a, we will now subsitute the value of the radius into the volume equation. This step should end up giving us an equation of the volume in terms of height.

 

Now that we have our volume equation in terms of h, we can now start off by following the same steps in part b as well. The first problem that some may come into contact with is the fact that we aren't given a value of the height, unlike part b where it gave you a value for the radius. In this question, we are given a radius instead. The solution to this is actually quite simple. Remember back in part a when we found what the radius was equal to using similar triangles? Well, we are going to use that and substitute in the value of the radius given in this question so we can get our height.

 

Before we go any further, let's first list off what pieces of information we know. Its always handy to keep track of what we've got so we don't get lost and confused in the future.

 

We're now ready to start solving for the rate of change of the height, dh/dt. Like in part b, let's first implicitly differentiate our equation for volume.

 

Now we can substitute the known values into our implicitly differentiated equation and solve for dh/dt.

 

Question solved, and without a calculator! I hope the helpful tips I put throughout this solution helped you a bit and guided you one step closer to not relying on the handy dandy calculator and gave you some better understanding with related rates. Till next time (Will there even be a next time? T__T)!

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