# Related Rates Problem

#### Hckyplayer8

##### Junior Member
First I would like to say thank you all who have helped myself and others on here. I was able to get through Exam 1 with a 78% which I will take considering math is my weakest subject and I'm taking Calc I online.

This week it is Related Rates which I'm taking nice and slow. With that, here is problem number 1.

1) A particle on the x-axis is moving to the right at 2 units per second. At a certain instant it is at the point (5,0). How rapidly is the distance between the particle and the point (0,9) on the y-axis changing at that point?

I drew my X-Y graph and marked points (0,9) and (5,0). I labeled X= distance and Y=time. The problem gives me dx/dt=2 units/sec.

The basic relationship looks like it should be a right triangle/Pythagorean Theorem.

So that is about where I am at because I know the next step is to sub the information into the equation but it's suppose to use implicit differentiation which I thought requires an unknown Y value...but it looks like I know the sides of the triangle A ^2+ B^2 but need C^2.

What am I missing?

#### lev888

##### Full Member
What's the equation of the distance from the particle to (0,9)? It's been a while, but as far as I remember if you have d = D(t), differentiating it would give you v = V(t), then you can plug in t at point (5,0) and you'll get the answer.

#### JeffM

##### Elite Member
Name things.

$$\displaystyle w = \text {distance between } (x,\ 0) \text { and } (0,\ 9).$$

$$\displaystyle w = \sqrt{(x - 0)^2 + (y - 9)^2} = \sqrt{x^2 + y^2 - 18y + 81}.$$

I presume you got to there. But the point is moving along the x-axis so the y coordinate is always 0.

Can you finish it up now?

#### Hckyplayer8

##### Junior Member
What's the equation of the distance from the particle to (0,9)? It's been a while, but as far as I remember if you have d = D(t), differentiating it would give you v = V(t), then you can plug in t at point (5,0) and you'll get the answer.
The equation isn't given. Believe it is up to me to figure out what the problem is asking and apply the proper equation.

Name things.

$$\displaystyle w = \text {distance between } (x,\ 0) \text { and } (0,\ 9).$$

$$\displaystyle w = \sqrt{(x - 0)^2 + (y - 9)^2} = \sqrt{x^2 + y^2 - 18y + 81}.$$

I presume you got to there. But the point is moving along the x-axis so the y coordinate is always 0.

Can you finish it up now?
Oh my god. It's an interval. Not a point. Doh! Between both of your posts I can see my original Pythagorean Theorem idea needs tossed.

Wait, if its an interval wouldn't (5,0) mean the particle is moving left? (X1, X2) with X1 being the starting point?

#### lev888

##### Full Member
The equation isn't given. Believe it is up to me to figure out what the problem is asking and apply the proper equation.
Oh my god. It's an interval. Not a point. Doh! Between both of your posts I can see my original Pythagorean Theorem idea needs tossed.
Wait, if its an interval wouldn't (5,0) mean the particle is moving left? (X1, X2) with X1 being the starting point?
You need to derive the equation.
Not sure what interval you are talking about.
What "needs tossed"?
It's moving to the right per problem statement.

#### JeffM

##### Elite Member
I think you are getting scrambled by trying to over-interpret things geometrically.

The particle has position (x, 0) so its y coordinate is always zero.

The distance w between the particle and the point (0, 9) is determined by the Pythagorean Theorem. The distance is that of the hypotenuse of the right triangle formed by (0, 9), (0, 0), and (x, 0). That was not an error on your part. You just forgot that y was a constant of 0.

We are told that the particle is moving to the right at 2 units per second. Ignoring units, that means

$$\displaystyle \dfrac{dx}{dt} = 2.$$

You know that w is a function of x. You want to know

$$\displaystyle \dfrac{dw}{dt} = \dfrac{dw}{dx} * \dfrac{dx}{dt}.$$

Once reduced to calculus, it essentially is a simple chain rule problem. Now when you work that out, it will say that the rate at which the distance is changing is some function of x. When x = 5, what is the numeric value of that function. Don't forget to put units on your final answer.

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#### HallsofIvy

##### Elite Member
Wait, if its an interval wouldn't (5,0) mean the particle is moving left? (X1, X2) with X1 being the starting point?
No, in your first post, you specifically said, "A particle on the x-axis is moving to the right at 2 units per second." Since it is on the x-axis, its y coordinate is always 0. Since it started at (5, 0) and is "moving to the right at 2 units per second", after t seconds it will be at (5+ 2t, 0). Use the Pythagorean theorem to find the straight line distance between (5+ 2t, 0) and (0, 9).

#### Hckyplayer8

##### Junior Member
Before I plug in my x coord I need to implicitly differentiate the square root of x2+y2-18y+81 correct?

So outside function is the square root and the inside is the x2+y2-18y+81.

Chain ruling it together it should be (1/2)x2+y2-18y+81(-1/2) times 2x+2y-18.

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#### JeffM

##### Elite Member
Before I plug in my x coord I need to implicitly differentiate the square root of x2+y2-18y+81 correct?

So outside function is the square root and the inside is the x2+y2-18y+81.

Chain ruling it together it should be (1/2)x2+y2-18y+81(-1/2) times 2x+2y-18.
The y coordinate of the moving particle is ALWAYS zero in this problem because the particle on the x-axis is moving to the right.

$$\displaystyle \therefore \sqrt{x^2 + y^2 - 18y + 81} = \sqrt{x^2 + 0^2 - 18(0) + 81} = \sqrt{x^2 + 81}.$$

#### Hckyplayer8

##### Junior Member
The y coordinate of the moving particle is ALWAYS zero in this problem because the particle on the x-axis is moving to the right.

$$\displaystyle \therefore \sqrt{x^2 + y^2 - 18y + 81} = \sqrt{x^2 + 0^2 - 18(0) + 81} = \sqrt{x^2 + 81}.$$
Oh. Since the y coordinate is a constant we know the y coordinate which makes it explicit form? But they also give us the x coordinate. So why are we leaving variable x in the equation instead of subbing 5 in?

#### JeffM

##### Elite Member
The question asks for the instantaneous rate of change in the distance between the particle and the point of (0, 9) when the particle is at position (5, 0).

Thus, x and y represent positions on the plane, not time and not distance. I missed this error that you made in your first post because the problem seemed "obvious" to me after 60 years of familiarity with calculus. (What is obvious after years of using calculus and what is obvious after a few hours of studying it is the difficulty in teaching anything.) So I apologize.

After what has been said before, you should recognize that the distance from (0, 9) to (x, 0), the location of the particle, which I have labeled w, is

$$\displaystyle w = \sqrt{(x - 0)^2 + (0 - 9)^2} = \sqrt{x^2 + 81}.$$

As you recognized from the start, this distance formula comes straight from the Pythagorean Theorem. With me so far?

You are correct that you can find w when x = 5 simply by plugging 5 into that distance equation:

$$\displaystyle x = 5 \implies w = \sqrt{5^2 + 81} = \sqrt{106}.$$

But we are not interested in w, which is a distance, but rather in the instantaneous rate of change in distance, namely

$$\displaystyle \dfrac{dw}{dt}.$$

Still with me? So the equation for w does not help us, at least not directly.

We know w is a function of x so if we knew $$\displaystyle \dfrac{dx}{dt}$$,

we could use the chain rule to calculate $$\displaystyle \dfrac{dw}{dt}$$

and then plug x = 5 into THAT equation. But we DO know $$\displaystyle \dfrac{dx}{dt}.$$

What is it? Can you finish the problem now? More importantly, do you understand the problem now?

#### Hckyplayer8

##### Junior Member
The question asks for the instantaneous rate of change in the distance between the particle and the point of (0, 9) when the particle is at position (5, 0).

Thus, x and y represent positions on the plane, not time and not distance. I missed this error that you made in your first post because the problem seemed "obvious" to me after 60 years of familiarity with calculus. (What is obvious after years of using calculus and what is obvious after a few hours of studying it is the difficulty in teaching anything.) So I apologize.

After what has been said before, you should recognize that the distance from (0, 9) to (x, 0), the location of the particle, which I have labeled w, is

$$\displaystyle w = \sqrt{(x - 0)^2 + (0 - 9)^2} = \sqrt{x^2 + 81}.$$

As you recognized from the start, this distance formula comes straight from the Pythagorean Theorem. With me so far?

You are correct that you can find w when x = 5 simply by plugging 5 into that distance equation:

$$\displaystyle x = 5 \implies w = \sqrt{5^2 + 81} = \sqrt{106}.$$

But we are not interested in w, which is a distance, but rather in the instantaneous rate of change in distance, namely

$$\displaystyle \dfrac{dw}{dt}.$$

Still with me? So the equation for w does not help us, at least not directly.

We know w is a function of x so if we knew $$\displaystyle \dfrac{dx}{dt}$$,

we could use the chain rule to calculate $$\displaystyle \dfrac{dw}{dt}$$

and then plug x = 5 into THAT equation. But we DO know $$\displaystyle \dfrac{dx}{dt}.$$

What is it? Can you finish the problem now? More importantly, do you understand the problem now?
No need to apologize. I appreciate you and everyone else taking time to help people like me out. What is really frustrating is being an above average student in all topics sans math. Learning this stuff...it's like trying to hit my way out of a wet paper bag.

The problem ultimately comes down to finding $$\displaystyle \dfrac{dw}{dt}$$ which reads as the derivative of w (the hypotenuse of the triangle) in regards to time. MATH]\dfrac{dx}{dt}[/MATH] reads as the derivative of coordinate x in regards to time which is given information. Thus, 2 units/sec.

Function 1 is the equation we created for length of w. Function 2 is time. So the final problem before plugging x = 5 is now

(1/2) x2+81(-1/2) times 2x +2

where I differentiated function 1 which needed the chain rule and then added the derivative of time which was already given in the problem.

Is that correct?

#### JeffM

##### Elite Member
The problem ultimately comes down to finding $$\displaystyle \dfrac{dw}{dt}$$ which reads as the derivative of w (the hypotenuse of the triangle) in regards to time.
This is absolutely correct.

It is frequently helpful in math to start by figuring out where you want to end up and then figuring out in reverse how to get there. Here are asked to find the rate of change with respect to time of the distance between a fixed location and a moving particle when that particle is at a specific location. That is where you want to end up. Translate that endpoint into the language of calculus. For that we need names for our variables. I arbitrarily chose w for distance and t for time. And the rate of change is

$$\displaystyle \dfrac{dw}{dt}.$$

Now if you just had a function relating t and w, you would differentiate and virtually be done. But there does not seem to be such a function directly. What functional relationships can you see? Well, there must be a relationship between the distance between the location of the fixed point and the location of the moving particle, If we could find that out and the relationship between the location of the moving point and time, we could differentiate both and use the chain rule. But wait, we already know the rate of change of the moving point with respect to time.

$$\displaystyle \dfrac{dx}{dt}$$ reads as the derivative of coordinate x in regards to time which is given information. Thus, 2 units/sec.
Correct. Very good. So now we have our plan: figure out the function relating the distance between the locations of the fixed point and the moving particle. Differentiate that, apply the chain rule to get the derivative we want, and plug in the appropriate values.

Function 1 is the equation we created for length of w. Function 2 is time.
The first sentence is correct. The second is not. The second function is not time, but the velocity of the moving particle as evidenced by its units: units of distance per second. The second function is already a derivative. (All derivatives are functions.)

So the final problem before plugging x = 5 is now

(1/2) x2+81(-1/2) times 2x +2

where I differentiated function 1 which needed the chain rule and then added the derivative of time which was already given in the problem.
Now you have jumped the rails.

Yes, you use the the chain rule to differentiate function 1.

$$\displaystyle w = \sqrt{x^2 + 81} \text { and } u = x^2 + 81 \implies w = \sqrt{u} = u^{1/2} \text { and } \dfrac{du}{dx} = 2x \implies$$

$$\displaystyle \dfrac{dw}{du} = \dfrac{1}{2} * u^{(1/2)-1} = \dfrac{1}{2\sqrt{u}} = \dfrac{1}{2\sqrt{x^2 + 81}} \text { and }$$

$$\displaystyle \dfrac{dw}{dx} = \dfrac{dw}{du} * \dfrac{du}{dx} = \dfrac{1}{\cancel 2 \sqrt{x^2 + 81}} * \cancel 2x = \dfrac{x}{\sqrt{x^2 + 81}}.$$

So, except for the cancellations, you did that correctly. BUT

$$\displaystyle \dfrac{dw}{dx} + \dfrac{dx}{dt} \ne \dfrac{dw}{dt}.$$

You do not get meaningful derivatives by adding derivatives that are not with respect to the same thing.

You must apply the chain rule again. (Although the Leibniz notation is only an analogy under standard analysis, it is frequently a helpful analogy.)

$$\displaystyle \dfrac{dw}{dt} = \dfrac{dw}{dx} * \dfrac{dx}{dt} = \dfrac{x}{\sqrt{x^2 + 81}} * 2 = \dfrac{2x}{\sqrt{x^2 + 81}}.$$

Now you can plug in x = 5.

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#### Hckyplayer8

##### Junior Member
This is absolutely correct.

It is frequently helpful in math to start by figuring out where you want to end up and then figuring out in reverse how to get there. Here are asked to find the rate of change with respect to time of the distance between a fixed location and a moving particle when that particle is at a specific location. That is where you want to end up. Translate that endpoint into the language of calculus. For that we need names for our variables. I arbitrarily chose w for distance and t for time. And the rate of change is

$$\displaystyle \dfrac{dw}{dt}.$$

Now if you just had a function relating t and w, you would differentiate and virtually be done. But there does not seem to be such a function directly. What functional relationships can you see? Well, there must be a relationship between the distance between the location of the fixed point and the location of the moving particle, If we could find that out and the relationship between the location of the moving point and time, we could differentiate both and use the chain rule. But wait, we already know the rate of change of the moving point with respect to time.

Correct. Very good. So now we have our plan: figure out the function relating the distance between the locations of the fixed point and the moving particle. Differentiate that, apply the chain rule to get the derivative we want, and plug in the appropriate values.

The first sentence is correct. The second is not. The second function is not time, but the velocity of the moving particle as evidenced by its units: units of distance per second. The second function is already a derivative. (All derivatives are functions.)

Now you have jumped the rails.

Yes, you use the the chain rule to differentiate function 1.

$$\displaystyle w = \sqrt{x^2 + 81} \text { and } u = x^2 + 81 \implies w = \sqrt{u} = u^{1/2} \text { and } \dfrac{du}{dx} = 2x \implies$$

$$\displaystyle \dfrac{dw}{du} = \dfrac{1}{2} * u^{(1/2)-1} = \dfrac{1}{2\sqrt{u}} = \dfrac{1}{2\sqrt{x^2 + 81}} \text { and }$$

$$\displaystyle \dfrac{dw}{dx} = \dfrac{dw}{du} * \dfrac{du}{dx} = \dfrac{1}{\cancel 2 \sqrt{x^2 + 81}} * \cancel 2x = \dfrac{x}{\sqrt{x^2 + 81}}.$$

So, except for the cancellations, you did that correctly. BUT

$$\displaystyle \dfrac{dw}{dx} + \dfrac{dx}{dt} \ne \dfrac{dw}{dt}.$$

You do not get meaningful derivatives by adding derivatives that are not with respect to the same thing.

You must apply the chain rule again. (Although the Leibniz notation is only an analogy under standard analysis, it is frequently a helpful analogy.)

$$\displaystyle \dfrac{dw}{dt} = \dfrac{dw}{dx} * \dfrac{dx}{dt} = \dfrac{x}{\sqrt{x^2 + 81}} * 2 = \dfrac{2x}{\sqrt{x^2 + 81}}.$$

Now you can plug in x = 5.
A recap just to make sure I am tracking...

By virtue of the particle moving to the right along the x-axis and the secondary point being centered on X=0, one could compose a right triangle and figure out that we are interested in finding out the derivative of the hypotenuse in order to answer the question. Outside of a brief lapse in judgment, this much seemed to come intuitively to me.

In order to figure out the derivative of the hypotenuse, one needs to figure out which functions to chain together to obtain the final derivative. In this case, rate of change was a result of a relationship between distance and time where $$\displaystyle \dfrac{dw}{dt}.$$ is the derivative of distance in regard to time.

The problem gave one of the two functions where the instantaneous velocity of the particle was moving 2 units/sec. The secondary function was the Pythagorean Theorem.

The equation a2+b2=c2 was inserted with a and b being the difference squared of the x and y coordinates. I would like clarification of why this was done. Why couldn't we just take the lengths of the opposite and adjacent sides squared and use the derivative of that?

Regardless of my question, the equation itself gets inserted into a square root being the inverse of the square function. After FOIL-ing, the known constant Y was inserted into the equation. The chain rule was then performed on that equation 1 (function 1).

Then we had to multiply that equation by the instantaneous velocity function to figure out what we really wanted to know which was the instantaneous rate of change.We had to multiple essentially because "denominators" of the derivatives were not the same.

#### JeffM

##### Elite Member
A couple of points.

First, a purist adopting standard analysis may heve a conniption to hear you talking about the denominators of derivatives. It is obvious that Leibniz thought that way, but under modern standard analysis, the derivative is an operator rather than a fraction. That is why I called the Leibniz notation of a derivative an analogy. Nevertheless, the Leibniz notation is a useful analogy when dealing with the chain rule: the "numerators" and "denominators" "cancel." And the units will also cancel using dimensional analysis.

Your second paragraph is spot on.

Now I am a little befuddled by what you say thereafter. It may be right, but I cannot be sure because your language is not precise. The Pythagorean Theorem is not a function although a function is specified by it in certain circumstances. Talking about derivatives without talking about what they are derivatives with respect to is meaningless. And all along, you seem to have been tangled up in some geometric web.

Distance in the Cartesian plane is always based on the Pythagorean Theorem. The distance between (p, q) and (r, s) =

$$\displaystyle \sqrt{(p - r)^2 + (q - s)^2}$$

because the vertical and horizontal axes are defined as being at right angles to each other. If you do not "see" this intuitively, draw a few examples. The distance is a hypotenuse.

In this problem, it is particularly easy to apply the distance formula. The fixed point is at (0, 9), and the moving point is at (x, 0). So two sides of the triangle are the x and y axes, and the distance is the hyptenuse running from (0, 9) to (x, 0).

$$\displaystyle \text {The distance from the fixed to the moving point} = w = \sqrt{(x - 0)^2 + (0 - 9)^2} = \sqrt{x^2 + 81}.$$

In functional notation, $$\displaystyle w = f(x).$$

I cannot figure out where that causes you perplexity.

Now letting t = time, we are told that $$\displaystyle \dfrac{dx}{dt} = 2.$$

But that means that $$\displaystyle x = g(t) \implies w = f(g(t)).$$

And we are asked to find $$\displaystyle \dfrac{dw}{dt}.$$

$$\displaystyle w = f(g(t)) \implies \dfrac{dw}{dt} = \dfrac{dw}{dx} * \dfrac{dx}{dt} = 2 * \dfrac{dw}{dx} \text { by the chain rule.}$$

I really want to help, but I need you to tell me where you find my explanation obscure.

#### Harry_the_cat

##### Senior Member
Here's how I approach these problems.

1. Draw a diagram if appropriate. Your diagram will look like a right angled triangle with the apex at the fixed point (0, 9), the right angle at the origin and a (movable) point at (x, 0) where x is the horizontal distance from the origin. (This is not 5 yet!)

2. Define your variables: x = horiz distance from origin to point; w = distance from (0, 9) to point - ie the hypotenuse on your diagram; t = time in seconds.

3. In terms of these variables, state:
(a) what you have been given .. you have been given $$\displaystyle \frac{dx}{dt} = 2$$ units/second
(b) what you want .. you want $$\displaystyle \frac{dw}{dt}$$
4. Write a "chain rule" statement involving these two derivatives .... $$\displaystyle \frac{dw}{dt} = \frac{dw}{dx}*\frac{dx}{dt}$$
5. OK, so you've got $$\displaystyle \frac{dx}{dt} = 2$$ and you need $$\displaystyle \frac{dw}{dt}$$. So now you have to find the "missing derivative" ie $$\displaystyle \frac{dw}{dx}$$
6. Before you can find $$\displaystyle \frac{dw}{dx}$$ you need a relationship between w and x (preferably write w as a function of x to avoid implicit differentiation) ... by Pythagoras' theorem $$\displaystyle w = \sqrt{x^2+81}$$.

7. Find $$\displaystyle \frac{dw}{dx}$$ by differentiation.

8. Substitute into your chain rule from Step 4.

9 To finish it off, evaluate using x=5 as given in the question.[/tex]

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#### Hckyplayer8

##### Junior Member
A couple of points.

First, a purist adopting standard analysis may heve a conniption to hear you talking about the denominators of derivatives. It is obvious that Leibniz thought that way, but under modern standard analysis, the derivative is an operator rather than a fraction. That is why I called the Leibniz notation of a derivative an analogy. Nevertheless, the Leibniz notation is a useful analogy when dealing with the chain rule: the "numerators" and "denominators" "cancel." And the units will also cancel using dimensional analysis.

Your second paragraph is spot on.

Now I am a little befuddled by what you say thereafter. It may be right, but I cannot be sure because your language is not precise. The Pythagorean Theorem is not a function although a function is specified by it in certain circumstances. Talking about derivatives without talking about what they are derivatives with respect to is meaningless. And all along, you seem to have been tangled up in some geometric web.

Distance in the Cartesian plane is always based on the Pythagorean Theorem. The distance between (p, q) and (r, s) =

$$\displaystyle \sqrt{(p - r)^2 + (q - s)^2}$$

because the vertical and horizontal axes are defined as being at right angles to each other. If you do not "see" this intuitively, draw a few examples. The distance is a hypotenuse.

In this problem, it is particularly easy to apply the distance formula. The fixed point is at (0, 9), and the moving point is at (x, 0). So two sides of the triangle are the x and y axes, and the distance is the hyptenuse running from (0, 9) to (x, 0).

$$\displaystyle \text {The distance from the fixed to the moving point} = w = \sqrt{(x - 0)^2 + (0 - 9)^2} = \sqrt{x^2 + 81}.$$

In functional notation, $$\displaystyle w = f(x).$$

I cannot figure out where that causes you perplexity.

Now letting t = time, we are told that $$\displaystyle \dfrac{dx}{dt} = 2.$$

But that means that $$\displaystyle x = g(t) \implies w = f(g(t)).$$

And we are asked to find $$\displaystyle \dfrac{dw}{dt}.$$

$$\displaystyle w = f(g(t)) \implies \dfrac{dw}{dt} = \dfrac{dw}{dx} * \dfrac{dx}{dt} = 2 * \dfrac{dw}{dx} \text { by the chain rule.}$$

I really want to help, but I need you to tell me where you find my explanation obscure.
Ah I get it now. I was trying to apply the formula to the earlier teachings of the class and just forgot that was literally the distance equation.

That said, forgive my ignorance but my question still stands. If we are trying to find the length of the hypotenuse and then subject that differentiation, why do we need the distance formula. I understand that is the correct formula to use but why can't we just treat it like a common right triangle problem?

For example a 3,4,5 triangle. 32+42=52

In this case 92+52=c2 and then apply the power rule to both sides.

Regardless of all of that, I feel confident that I can replicate what has been taught to me in here and I appreciate the time.

#### lev888

##### Full Member
That said, forgive my ignorance but my question still stands. If we are trying to find the length of the hypotenuse and then subject that differentiation, why do we need the distance formula. I understand that is the correct formula to use but why can't we just treat it like a common right triangle problem?
You can find the distance at a particular point. But then what do you do with it? You can't differentiate it, it's too late - it's just a number.
You are asked to find the rate of change of a function at a point. This is done by differentiating the function and plugging in the parameter corresponding to that point.

#### Harry_the_cat

##### Senior Member
The distance formula is the same as Pythagoras' theorem for right angled triangles. That's where the distance formula comes from.

#### JeffM

##### Elite Member
I think lev answered your question. To find the distance from (0, 9) to (5, 0), you can plug the specific numbers for your problem into the Pythagorean Theorem (or its corollary the distance formula) and chug away arithmetically. But you are not asked to do that; you are asked to find the rate of change of the distance. Before you can plug any numbers and chug away arithmetically, you must first find a formula for the rate of change to plug into, and that takes calculus in addition to the geometry of the problem.

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