Why doesn't this work? (using sine instead of tangent)

[allegansveritatem]

Here is the problem, actually it is an example from my text book:

I worked this out as it is shown here and got the same result as the book. Then I began wondering
what would happen if I used the sine function instead of the tangent for the equation
and I got this result:

Why didn't it come out the same? I mean shouldn't I have gotten the same result whether I
x was equated in terms of tan or in terms of sine? Or did I make some stupid little mistake
that I am blind to and have been all day, I might add? I have redone this sine version of the problem 5 or 6 times
always with the same outcome.

 

[Dr.Peterson]

Here is the problem, actually it is an example from my text book:

I worked this out as it is shown here and got the same result as the book. Then I began wondering
what would happen if I used the sine function instead of the tangent for the equation
and I got this result:

Why didn't it come out the same? I mean shouldn't I have gotten the same result whether I
x was equated in terms of tan or in terms of sine? Or did I make some stupid little mistake
that I am blind to and have been all day, I might add? I have redone this sine version of the problem 5 or 6 times
always with the same outcome.

Yes, you made a stupid mistake. (You asked.) It's a common one when people are first learning about related rates.

The hypotenuse is 25 at the moment they are asking about; it is not 25 while he walks. So you can't differentiate [imath]x = 25\sin(\theta)[/imath] and expect to get the correct rates.

Specifically, when you make your diagram like this,

​

it is as if you told the man to stand still while you make a movie of him! To find the rate of the searchlight, you have to set him free to move. Remove that 25-foot chain.

The correct procedure is to write an equation that describes the relationship between the variables at all times, then differentiate it, and only use the information about one specific moment after you have differentiated.

 

[allegansveritatem]

Yes, you made a stupid mistake. (You asked.) It's a common one when people are first learning about related rates.

The hypotenuse is 25 at the moment they are asking about; it is not 25 while he walks. So you can't differentiate [imath]x = 25\sin(\theta)[/imath] and expect to get the correct rates.

Specifically, when you make your diagram like this,

View attachment 35413​

it is as if you told the man to stand still while you make a movie of him! To find the rate of the searchlight, you have to set him free to move. Remove that 25-foot chain.

The correct procedure is to write an equation that describes the relationship between the variables at all times, then differentiate it, and only use the information about one specific moment after you have differentiated.

I see that you are right. I have to go back and rethink this whole matter of related rates. Yes, absolutely, some of those parameters are dynamic. I was not thinking the thing through. Thanks.

 

[R.M.]

Three important tips for related rates problems:
1. Fill in these blanks: find __________ when _____________. In your example, find d[imath]\theta[/imath]/dt when x = 15.
2. Write the equation ABSTRACTLY first.
3. Differentiate. Because you did not substitute anything yet, you don't have any terms zero out.

Now substitute in your values, and you are on your way.

 

[allegansveritatem]

Well, I went at it again today with this, keeping in mind that there are certain dynamisms in this business of related rates that have to be kept in mind. But I soon found myself walking in darkness again. Here is what I did:

I first found the derivative formula and then plugged in the values and came up with the same result as before, i.e., .2.
So I am still not out of the woods. I still don't know why the author of this book, James Stewart, chose tan, which entails a squared derivative and adds thus another layer of complexity instead of the sine. I thought maybe I could clarify the matter and maybe find my mistake, if any, by solving the problem with both tan and sine, side by side. And, as the image below shows, nothing came of it:

So....what's up?

 

[Dr.Peterson]

I still don't know why the author of this book, James Stewart, chose tan, which entails a squared derivative and adds thus another layer of complexity instead of the sine.

He chose to use the tangent because only that reflects the actual relationship of the variables while the man walks!

Your equation (and the way you differentiated it) says that h is constant, rather than that the distance of 20 is constant. But, as I said, there is no 25-foot chain from the searchlight to the man! Since h varies, what you did is wrong. (You'd have to use the product rule, and know dh/dt, when you differentiate.)

Whether the equation is correct is far more important than how hard the differentiation is. Do the right thing, however hard, rather than do the wrong thing because it's easy. (Do you always look for lost objects under the street light?)

 

[allegansveritatem]

He chose to use the tangent because only that reflects the actual relationship of the variables while the man walks!

Your equation (and the way you differentiated it) says that h is constant, rather than that the distance of 20 is constant. But, as I said, there is no 25-foot chain from the searchlight to the man! Since h varies, what you did is wrong. (You'd have to use the product rule, and know dh/dt, when you differentiate.)

Whether the equation is correct is far more important than how hard the differentiation is. Do the right thing, however hard, rather than do the wrong thing because it's easy. (Do you always look for lost objects under the street light?)

Thanks for your reply. I will have to wait til the morning to digest it--my brain shuts down after 10PM these days. I will get back with my results tomorrow.

 

[allegansveritatem]

He chose to use the tangent because only that reflects the actual relationship of the variables while the man walks!

Your equation (and the way you differentiated it) says that h is constant, rather than that the distance of 20 is constant. But, as I said, there is no 25-foot chain from the searchlight to the man! Since h varies, what you did is wrong. (You'd have to use the product rule, and know dh/dt, when you differentiate.)

Whether the equation is correct is far more important than how hard the differentiation is. Do the right thing, however hard, rather than do the wrong thing because it's easy. (Do you always look for lost objects under the street light?)

So, once again I went at it today, this time with better understanding and higher hopes. I Think I understand now why Dr. Stewart used the tangent instead of the sine function. As Dr. Peterson explained, the tangent (sine/cosine) is the function that allows all the information given in the problem to be accessed. You have the 20 feet, the 15 feet and the hypotenuse too. Anyway, here is my final shot at the thing, using the product rule to find the derivative. I came out 3 thousandths of a point less than the amount I would have gotten had I used the method in the text, but, hey!, that ain't hardly anything to complain about...although I would like to know exactly why this is so.

 

[Dr.Peterson]

So, once again I went at it today, this time with better understanding and higher hopes. I Think I understand now why Dr. Stewart used the tangent instead of the sine function. As Dr. Peterson explained, the tangent (sine/cosine) is the function that allows all the information given in the problem to be accessed. You have the 20 feet, the 15 feet and the hypotenuse too. Anyway, here is my final shot at the thing, using the product rule to find the derivative. I came out 3 thousandths of a point less than the amount I would have gotten had I used the method in the text, but, hey!, that ain't hardly anything to complain about...although I would like to know exactly why this is so.
View attachment 35472

One mistake in your work is here:



You are differentiating implicitly, so you need to treat y and [imath]\theta[/imath] as functions of t, and use the chain rule. So you need dy/dt after the first term, and d[imath]\theta[/imath]/dt after the secpnd term.

As a result of this mistake, the rest of the work is nonsense. You can't solve for d[imath]\theta[/imath]/dt when it isn't even in the equation you are solving. The fact that your answer is close to the correct one is pure coincidence.

On the other hand, since y is a constant, 20, dy/dt = 0, and you didn't really need the product rule.

 

[Steven G]

Why Stewart and I would use tan instead of sin. The problem only gave you the opposite and adjacent sides of the right triangle. Possibly in computing the hypotenuse you made a mistake and now will make another mistake by using the wrong hypotenuse while computing the sine.

It is customary to advise students to try not to use their own calculations when calculating other values. Sometimes you can't avoid using your calculations.


In your case, you find the unknown values too early on and that messes you up. If I was you, my last step would have been finding h.
In this way you would have avoided thinking that h=25 since in your diagram it would have said h. This is a very big hint for you to have success with related rates problem.

 

[allegansveritatem]

One mistake in your work is here:

View attachment 35474

You are differentiating implicitly, so you need to treat y and [imath]\theta[/imath] as functions of t, and use the chain rule. So you need dy/dt after the first term, and d[imath]\theta[/imath]/dt after the secpnd term.

As a result of this mistake, the rest of the work is nonsense. You can't solve for d[imath]\theta[/imath]/dt when it isn't even in the equation you are solving. The fact that your answer is close to the correct one is pure coincidence.

On the other hand, since y is a constant, 20, dy/dt = 0, and you didn't really need the product rule.

Yes, I see that I lost track of that dy/dt. That should have been in there somewhere...I am confused here as to how implicit differentiation is to be accounted for here. I have to go back and go over that again. Then thee is the business of y=20. I will go over this stuff again today an d try to make better sense of it. Thanks for the pointers.

 

[allegansveritatem]

Why Stewart and I would use tan instead of sin. The problem only gave you the opposite and adjacent sides of the right triangle. Possibly in computing the hypotenuse you made a mistake and now will make another mistake by using the wrong hypotenuse while computing the sine.

It is customary to advise students to try not to use their own calculations when calculating other values. Sometimes you can't avoid using your calculations.


In your case, you find the unknown values too early on and that messes you up. If I was you, my last step would have been finding h.
In this way you would have avoided thinking that h=25 since in your diagram it would have said h. This is a very big hint for you to have success with related rates problem.

Thanks for the comment. I will think about what you have said.

 

[allegansveritatem]

I worked on this matter again yesterday. I went through the thing in pretty much the same way the author does. I used the information given in the problem: We are given that when the light is closest to the path along which the man is walking it is 20 feet from it. and we are asked , if said light is tracking the man, how fast is it going to be moving when the man, walking at 4 ft/sec, gets 15 feet away from the point where the path is closest to the light. So we have the rate of the walker and want the rate of the turning of the lamp. So we develop an equation that equates all this information in such a way that we can differentiate both sides with respect to time and come up with what we want to know. We use tan instead of sin because it relates the given parameters.

This example stopped me in the first place because I couldn't make out why the author used tan, whose
derivative is more unwieldy than sin's. Now I know--thanks to the contributors to this thread-- that using sin would have been getting ahead of ourselves.
If I thought anybody were listening I would advise the publishers--I think the author is deceased-- to insert a note to that effect in the next edition. But, maybe students at this level are expected to know things like that. Fond expectation, that.

 

[Steven G]

My concern is that your diagram does not show what the length of the hypotenuse is. Prior you made a mistake by putting a constant in the diagram for the hypotenuse and now you put nothing in for the hypotenuse.

 

[allegansveritatem]

My concern is that your diagram does not show what the length of the hypotenuse is. Prior you made a mistake by putting a constant in the diagram for the hypotenuse and now you put nothing in for the hypotenuse.

Right. I should have put an h there in as much as it is as yet unknown, at least it hasn't been given in the problem's description.

 

[Dr.Peterson]

My concern is that your diagram does not show what the length of the hypotenuse is. Prior you made a mistake by putting a constant in the diagram for the hypotenuse and now you put nothing in for the hypotenuse.

Right. I should have put an h there in as much as it is as yet unknown, at least it hasn't been given in the problem's description.

Why is it necessary to label h when it is not needed to solve the problem?

The only error is in calculating it without having labeled it, and without needing it. Leaving it unlabeled and unused would have been fine.

 

[Steven G]

Everything that you know should be in the diagram. Why? Because you don't know what you will need yet. What if the OP decided to use the sine function?

In the end, you are correct that h was not needed. And of course you tell your students only to label what is needed. This is just a personal attack on me and is at least the 2nd one