Related Rates Part III

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Hckyplayer8

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A particle on the x-axis is moving to the right at 2 units per second. A second particle is moving down the y axis at the rate of 3 units per second. At a certain instant, the first is at (5,0) and the second is at (0,7). How rapidly is the angle between the x-axis and the line joining the two particles changing at that instant? Are they moving towards or away from each other at that instant?

Like my first post, we have a problem involving a right triangle. This time it seems we are interested in angle theta that is made up of the hypotenuse and adjacent. We know the length of the opposite and adjacent sides which means we use the tangent equation which should be the first function no?

After that, I'm not sure what else to do.

d theta/dt = d tan/dt and ?
 
I would begin with:

[MATH]\tan(\theta)=\frac{y}{x}\implies y=x\tan(\theta)[/MATH]
Hence:

[MATH]\d{y}{t}=\d{x}{t}\tan(\theta)+x\sec^2(\theta)\d{\theta}{t}[/MATH]
Now, solve for [MATH]\d{\theta}{t}[/MATH] and utilize a Pythagorean identity for the square of the secant function. What do you get?
 
MarkFL said:
I would begin with:

[MATH]\tan(\theta)=\frac{y}{x}\implies y=x\tan(\theta)[/MATH]
Hence:

[MATH]\d{y}{t}=\d{x}{t}\tan(\theta)+x\sec^2(\theta)\d{\theta}{t}[/MATH]
Now, solve for [MATH]\d{\theta}{t}[/MATH] and utilize a Pythagorean identity for the square of the secant function. What do you get?
Click to expand...

We know dy/dt = -3 units/sec and dx/dt=2 units/sec

So plugging that into the equation,

-3 = 2(tan [MATH] {\theta}[/MATH]) + x sec2[MATH] {\theta}[/MATH] d[MATH] {\theta}[/MATH]/dt

The next step is to differentiate?
 
No, I would first solve for [MATH]\d{\theta}{t}[/MATH]:

[MATH]\d{\theta}{t}=\frac{\d{y}{t}-\d{x}{t}\tan(\theta)}{x\sec^2(\theta)}[/MATH]
Apply Pythagorean identity:

[MATH]\d{\theta}{t}=\frac{\d{y}{t}-\d{x}{t}\tan(\theta)}{x\left(1+\tan^2(\theta)\right)}[/MATH]
Substitute for \(\tan(\theta)\):

[MATH]\d{\theta}{t}=\frac{\d{y}{t}-\d{x}{t}\frac{y}{x}}{x\left(1+\frac{y^2}{x^2}\right)}=\frac{x\d{y}{t}-y\d{x}{t}}{x^2+y^2}[/MATH]
Now, plug in the given data...
 
MarkFL said:
I would begin with:

[MATH]\tan(\theta)=\frac{y}{x}\implies y=x\tan(\theta)[/MATH]
Hence:

[MATH]\d{y}{t}=\d{x}{t}\tan(\theta)+x\sec^2(\theta)\d{\theta}{t}[/MATH]
Now, solve for [MATH]\d{\theta}{t}[/MATH] and utilize a Pythagorean identity for the square of the secant function. What do you get?
Click to expand...

Okay reverting back to your first reply and spelling it out in plain text just to make sure I am tracking...

The derivative of y in regard to time is equal to the derivative of x in regard to time times tangent of theta plus variable x secant squared of theta times the derivative of theta in regards to time. So you already differentiated by using the product rule. I understand all of that sans the last part. Where did the derivative of theta in regards to time come from?
 
Another (easier?) approach after you get to

13273
is to sub in the values you know (seeing you have now done the differentiation).

You know x, y, dx/dt, dy/dt. You can calculate theta or better still, tan(theta) and sec(theta) from a simple diagram when x=5 and y=7.
Your only unknown is d(theta)/dt which is what you want.
 
Harry_the_cat said:
Another (easier?) approach after you get to

View attachment 13273
is to sub in the values you know (seeing you have now done the differentiation).

You know x, y, dx/dt, dy/dt. You can calculate theta or better still, tan(theta) and sec(theta) from a simple diagram when x=5 and y=7.
Your only unknown is d(theta)/dt which is what you want.
Click to expand...

That's what I was trying to do in post 3.
 
Hckyplayer8 said:
Because the derivative of theta in regards to time is based off the derivative of the tangent of theta which is based of the derivative of the moving particles?
Click to expand...
Not sure I can follow what you are trying to say but have a look below:

\(\displaystyle y=x * tan\theta\)

\(\displaystyle \frac{dy}{dt} =\frac{d}{dt}(x*tan\theta)\)
\(\displaystyle =x* \frac{d}{dt}(tan\theta) + \frac{dx}{dt}* tan\theta\) ….. product rule
\(\displaystyle =x* \frac{d}{d\theta}(tan\theta)*\frac{d\theta}{dt} + \frac{dx}{dt}* tan\theta\) ….. chain rule applied to \(\displaystyle \frac{d}{dt}(tan\theta)\)
\(\displaystyle =x* sec^2 \theta*\frac{d\theta}{dt} + \frac{dx}{dt}* tan\theta\)
 
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Harry_the_cat said:
Not sure I can follow what you are trying to say but have a look below:

\(\displaystyle y=x * tan\theta\)

\(\displaystyle \frac{dy}{dt} =\frac{d}{dt}(x*tan\theta)\)
\(\displaystyle =x* \frac{d}{dt}(tan\theta) + \frac{dx}{dt}* tan\theta\) ….. product rule
\(\displaystyle =x* \frac{d}{d\theta}(tan\theta)*\frac{d\theta}{dt} + \frac{dx}{dt}* tan\theta\) ….. chain rule applied to \(\displaystyle \frac{d}{dt}(tan\theta)\)
\(\displaystyle =x* sec^2 \theta*\frac{d\theta}{dt} + \frac{dx}{dt}* tan\theta\)
Click to expand...

I'm with you through the product rule. What would the chain rule look like in a different, non-Leibniz notation?

For example, [(2x+1)7]
Function 1 = 2x+1
Function 2 = A base raised to the 7th power

Outside function evaluated at the inside function times the derivative of the inside function.

So 7(2x+1)6 * 2 = 14(2x+1)6
 
Hckyplayer8 said:
A particle on the x-axis is moving to the right at 2 units per second. A second particle is moving down the y axis at the rate of 3 units per second. At a certain instant, the first is at (5,0) and the second is at (0,7). How rapidly is the angle between the x-axis and the line joining the two particles changing at that instant? Are they moving towards or away from each other at that instant?

Like my first post, we have a problem involving a right triangle. This time it seems we are interested in angle theta that is made up of the hypotenuse and adjacent. We know the length of the opposite and adjacent sides which means we use the tangent equation which should be the first function no?

After that, I'm not sure what else to do.

d theta/dt = d tan/dt and ?
Click to expand...
The tangent function may look like a good starting point, but I would start with the fact that to find the derivative of theta with respect to time t, I need a function relating theta and t. And I will need functions relating other variables to t. How do I get things into terms of t?

[MATH]\theta = \text {some function.}[/MATH]
I am interested in f'(t) at the time when x = 5 and y = 7. Let's say that occurs when t = 0. (This is totally arbitrary except it will make computations simple.)

[MATH]\therefore x = 5 + 2t \text { and } y = 7 - 3t \implies \dfrac{dx}{dt} = 2 \text { and } \dfrac{dy}{dt} = -\ 3.[/MATH]
Does that make sense?

Now I can find a relation among x, y, and a function of theta as you saw right away. But that does not directly give

[MATH]\theta = \text {some function.}[/MATH] What can we do? Use the inverse.

[MATH]tan ( \theta ) = \dfrac{y}{x} \implies \theta = arctan \left ( \dfrac{x}{y} \right ).[/MATH]
[MATH]u = \dfrac{y}{x} \implies \theta = arctan(u) \implies \dfrac{d \theta }{du} = \dfrac{1}{1 + u^2} = \dfrac{1}{1 + \dfrac{y^2}{x^2}} = \dfrac{x^2}{x^2 + y^2}.[/MATH]
[MATH]\text {Also, } u = \dfrac{y}{x} \implies \dfrac{du}{dt} = \dfrac{x * \dfrac{dy}{dt} - y * \dfrac{dx}{dt}}{x^2} = \dfrac{-\ 3x - 2y}{x^2}.[/MATH]
Now we are set to go with the chain rule.

[MATH]\dfrac{d \theta }{dt} = \dfrac{d \theta}{du} * \dfrac{du}{dt} = \dfrac{x^2}{x^2 + y^2} * \dfrac{-\ 3x - 2y}{x^2} = \dfrac{-\ 3x - 2y}{x^2 + y^2}.[/MATH]
Now we carefully set up that at t = 0, x = 5 and y = 7. So

[MATH]\therefore t = 0 \implies \dfrac{d \theta }{dt} = \dfrac{-\ 3(5) - 2(7)}{5^2} = -\ \dfrac{29}{25}.[/MATH]
Does the sign make sense? Why? What units apply to the answer?
 
Oops

[MATH]t = 0 \implies \dfrac{d \theta}{dt} = \dfrac{-\ 3(5) - 2(7)}{5^2 + 7^2} = -\ \dfrac{15 + 14}{25 + 49} = -\ \dfrac{29}{74}.[/MATH]
 
Hckyplayer8 said:
I'm with you through the product rule. What would the chain rule look like in a different, non-Leibniz notation?

For example, [(2x+1)7]
Function 1 = 2x+1
Function 2 = A base raised to the 7th power

Outside function evaluated at the inside function times the derivative of the inside function.

So 7(2x+1)6 * 2 = 14(2x+1)6
Click to expand...
I prefer this notation.
The chain rule states that \(\displaystyle \frac{dy}{dt}= \frac{dy}{d\theta}*\frac{d\theta}{dt}\).
If \(\displaystyle y = tan\theta\), this becomes:
\(\displaystyle \frac{d}{dt}(tan\theta)=\frac{d}{d\theta}(tan\theta)*\frac{d\theta}{dt}\)

In alternative notation:
If \(\displaystyle y = f(\theta) \) and \(\displaystyle \theta= g(t)\), then \(\displaystyle f'(t) = f'(\theta)*g'(t)\).
 
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Looking at this entire thread in the light of a new day, I have a few supplementary comments.

First, I did not mean to imply even a suspicion that Mark's method is invalid. It is fine, every bit as valid as mine. I myself just find it more intuitive to address the question of what is the derivative of u with respect to v by starting with u = f(v) rather than with v = g(u). In this case at least, it makes the resulting math somewhat easier.

Second, I probably should have pointed out that my method will indeed lead to theta as a function of time. I get

[MATH]\theta = arctan \left ( \dfrac{y}{x} \right ) = arctan \left ( \dfrac{7 - 3t}{5 + 2t} \right ).[/MATH]
That is a bit messy to deal with algebraically so I did a u-substitution without saying so. That was less than clear.

Third, I found another typographical error in my first post.

[MATH]tan ( \theta ) = \dfrac{y}{x} \implies \theta = arctan \dfrac{y}{x}[/MATH]
is what I meant and should have said.
 
JeffM said:
The tangent function may look like a good starting point, but I would start with the fact that to find the derivative of theta with respect to time t, I need a function relating theta and t. And I will need functions relating other variables to t. How do I get things into terms of t?

[MATH]\theta = \text {some function.}[/MATH]
I am interested in f'(t) at the time when x = 5 and y = 7. Let's say that occurs when t = 0. (This is totally arbitrary except it will make computations simple.)

[MATH]\therefore x = 5 + 2t \text { and } y = 7 - 3t \implies \dfrac{dx}{dt} = 2 \text { and } \dfrac{dy}{dt} = -\ 3.[/MATH]
Does that make sense?
Click to expand...

Yes. Similar to the the algebraic formula of a derivative, we are figuring out what happens as H approaches 0 of the equation F(a+h) - f(a) over h.

JeffM said:
Now I can find a relation among x, y, and a function of theta as you saw right away. But that does not directly give

[MATH]\theta = \text {some function.}[/MATH] What can we do? Use the inverse.

[MATH]tan ( \theta ) = \dfrac{y}{x} \implies \theta = arctan \left ( \dfrac{x}{y} \right ).[/MATH]
[MATH]u = \dfrac{y}{x} \implies \theta = arctan(u) \implies \dfrac{d \theta }{du} = \dfrac{1}{1 + u^2} = \dfrac{1}{1 + \dfrac{y^2}{x^2}} = \dfrac{x^2}{x^2 + y^2}.[/MATH]
[MATH]\text {Also, } u = \dfrac{y}{x} \implies \dfrac{du}{dt} = \dfrac{x * \dfrac{dy}{dt} - y * \dfrac{dx}{dt}}{x^2} = \dfrac{-\ 3x - 2y}{x^2}.[/MATH]
Click to expand...

Is u just your chosen placeholder for the inside function of the composition that is tan (inside function)? In another post, you mention substituting u in but I just wanted to be sure.

JeffM said:
Now we are set to go with the chain rule.

[MATH]\dfrac{d \theta }{dt} = \dfrac{d \theta}{du} * \dfrac{du}{dt} = \dfrac{x^2}{x^2 + y^2} * \dfrac{-\ 3x - 2y}{x^2} = \dfrac{-\ 3x - 2y}{x^2 + y^2}.[/MATH]
Now we carefully set up that at t = 0, x = 5 and y = 7. So

[MATH]\therefore t = 0 \implies \dfrac{d \theta }{dt} = \dfrac{-\ 3(5) - 2(7)}{5^2} = -\ \dfrac{29}{25}.[/MATH]
Does the sign make sense? Why? What units apply to the answer?
Click to expand...

The sign? The units should be radians since we are deriving an angle. Radians per second based off the problem.
 
For part B, are the particles moving towards or away from each other...intuitively I would say moving towards one another by virtue of y's rate of change being larger than x and this would be valid up until y crosses the origin. If that is correct, how can I prove it mathematically?
 
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Hckyplayer8 said:
For part B, are the particles moving towards or away from each other...intuitively I would say moving towards one another by virtue of y's rate of change being larger than x and this would be valid up until y crosses the origin. If that is correct, how can I prove it mathematically?
Click to expand...
How do we determine whether a function is increasing or decreasing at a point?
 
Couple of responses.

Yes, I substituted u for y/x. It turned out not to be much of a convenience in this case. But I tend to replace expressions with simple variables to avoid complex calculation for as long as possible and to avoid mistakes in transcribing expressions. It is never necessary, but it reduces the number of dumb mistakes I make.

You are right about the units being radians per second. The nice clean derivatives of the trigonometric functions and their inverses assume that angles are measured in radians.

I asked about the sign of [MATH]\dfrac{d \theta }{dt}.[/MATH] for a reason. It is negative; therefore the angle is decreasing when t = 0. If you think about what the math is describing, you will see that such a qualitative result makes sense. It is always a good idea to ask youself whether your mathematical results make qualitative sense. When they don't, it suggests looking for an error in your math.
 
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