A point moving in the plane

[Miszka]

Hey! I'm having troubles with that exercise from the Calculus book:

A point in the plane moves in such a way that it is always twice as far from (0,0) as it is from (0,1).

1. (a) Show that the point moves on a circle.
2. (b) At the moment when the point crosses the segment between (0,0) and (0, 1), what is dy/dt?
3. (c) Where is the point when dy/dt = dx/dt? (You may assume that dx/dt and dy/dt are not simultaneously zero.)

Could you help me with that task? It's in the chapter covering Related Rates and Parametric Curves. I have literally no idea how to do that, I did some previous exercises and had no bigger problems with them. I would be very grateful!

 

[Deleted member 4993]

Hey! I'm having troubles with that exercise from the Calculus book:

A point in the plane moves in such a way that it is always twice as far from (0,0) as it is from (0,1).

1. (a) Show that the point moves on a circle.
2. (b) At the moment when the point crosses the segment between (0,0) and (0, 1), what is dy/dt?
3. (c) Where is the point when dy/dt = dx/dt? (You may assume that dx/dt and dy/dt are not simultaneously zero.)

Could you help me with that task? It's in the chapter covering Related Rates and Parametric Curves. I have literally no idea how to do that, I did some previous exercises and had no bigger problems with them. I would be very grateful!

Let the origin be O (0,0) and the point be B (0,1)

Let the co-ordinate of the wandering point (P) be (x₁, y₁) at a given time.

What is the length of the line OP?

What is the length of the line PB?

continue.....

 

[pka]

A point in the plane moves in such a way that it is always twice as far from (0,0) as it is from (0,1).
1 (a) Show that the point moves on a circle.
1 (b) At the moment when the point crosses the segment between (0,0) and (0, 1), what is dy/dt?
1 (c) Where is the point when dy/dt = dx/dt? (You may assume that dx/dt and dy/dt are not simultaneously zero.)

Lets start with part (a) The distance the point is from \(\displaystyle (0,0)\) is \(\displaystyle \sqrt{x^2+y^2}\)
The distance the point is from \(\displaystyle (0,1)\) is \(\displaystyle \sqrt{x^2+(y-1)^2}\)
You now need to make an equation that expresses the fact the point is twice the distance from one as from the other.
Solve that showing the answer in standard circular form. Post your answer so that we can continue.

 

[Miszka]

Lets start with part (a) The distance the point is from \(\displaystyle (0,0)\) is \(\displaystyle \sqrt{x^2+y^2}\)
The distance the point is from \(\displaystyle (0,1)\) is \(\displaystyle \sqrt{x^2+(y-1)^2}\)
You now need to make an equation that expresses the fact the point is twice the distance from one as from the other.
Solve that showing the answer in standard circular form. Post your answer so that we can continue.

Alright, I did as you said and I got a circular form:
[MATH]3x^{2} + 3(y - \frac{4}{3})^{2} = \frac{4}{3}[/MATH]I believe that that's the answer to the first question, thank you. And how should I do the next point? I tried to differentiate that thing and got that:
[MATH]\frac{dy}{dt} = \frac{-6x}{6y - 8}[/MATH]And that's equal 0 because both points have x equal to 0? Am I right?

 

[pka]

b) the point The distance the point is from \(\displaystyle \left(0,\dfrac{1}{3}\right)\) is where intersects the segment (0,0) to (0,1).

 

[Miszka]

b) the point The distance the point is from \(\displaystyle \left(0,\dfrac{1}{3}\right)\) is where intersects the segment (0,0) to (0,1).

I'm sorry but I don't really understand what you mean. What's that point and how did we get that? And what should I do with that?

 

[Dr.Peterson]

I'm not sure what pka was referring to. I think your answer is close, but what you said is not very clear, and is not quite correct.

Did you notice that nothing is said about t in the problem? We have no idea how fast the point is moving at any given time, so in general we can say nothing about dy/dt. What you calculated is not dy/dt but dy/dx. You need to state clearly how it tells you what dy/dt is at that point. Similar thinking will be needed in part (c).

You could also answer both questions directly from the result of part (a). Sketch that circle and the given points, and think about the implications.

 

[JeffM]

OK

[MATH]\sqrt{(x - 0)^2 + (y - 0)^2} = 2 * \sqrt{(x - 0)^2 + (y - 1)^2} \implies x^2 + y^2 = 4x^2 + 4y^2 - 8y + 4 \implies[/MATH]
[MATH]3y^2 + 8y + 3x^2 = -\ 4 \implies y^2 + 2 * \dfrac{4}{3} * y + x^2 = -\ \dfrac{4}{3} \implies[/MATH]
[MATH]\left \{ y^2 + 2\left ( \dfrac{4}{3} \right )y + \left ( \dfrac{4}{3} \right )^2 \right\} + x^2 = \dfrac{16}{9} - \dfrac{4}{3} = \dfrac{4}{9} \implies [/MATH]
[MATH]\left ( y - \dfrac{4}{3} \right )^2 + (x - 0)^2 = \left ( \dfrac{2}{3} \right )^2.[/MATH]
So that looks good, but technically it does not define a differentiable function.

Why?

However, we are interested in a point above the x-axis, and the locus of points above the x-axis does correspond to the function

[MATH]y = \sqrt{\dfrac{4}{9} - x^2} + \dfrac{4}{3}.[/MATH]
So what is dy/dx there?

I agree with Dr. Peterson that questions about dy/dt seem to come from outer space. But the questions seem to ask about them. Is there a part of the problem that you did not give us?

 

[Steven G]

OK

[MATH]\sqrt{(x - 0)^2 + (y - 0)^2} = 2 * \sqrt{(x - 0)^2 + (y - 1)^2} \implies x^2 + y^2 = 4x^2 + 4y^2 - 8y + 4 \implies[/MATH]
[MATH]3y^2 + 8y + 3x^2 = -\ 4 \implies y^2 + 2 * \dfrac{4}{3} * y + x^2 = -\ \dfrac{4}{3} \implies[/MATH]
[MATH]\left \{ y^2 + 2\left ( \dfrac{4}{3} \right )y + \left ( \dfrac{4}{3} \right )^2 \right\} + x^2 = \dfrac{16}{9} - \dfrac{4}{3} = \dfrac{4}{9} \implies [/MATH]
[MATH]\left ( y - \dfrac{4}{3} \right )^2 + (x - 0)^2 = \left ( \dfrac{2}{3} \right )^2.[/MATH]
So that looks good, but technically it does not define a differentiable function.

Why?

However, we are interested in a point above the x-axis, and the locus of points above the x-axis does correspond to the function

[MATH]y = \sqrt{\dfrac{4}{9} - x^2} + \dfrac{4}{3}.[/MATH]
So what is dy/dx there?

I agree with Dr. Peterson that questions about dy/dt seem to come from outer space. But the questions seem to ask about them. Is there a part of the problem that you did not give us?

1st line has -8y while the 2nd line has + 8y. To the OP, this was a silly error that we all make from time to time (it seems to happen to me way too often). However at the end the correct answer was obtained (ie -8y became 8y and then back to -8y)

 

[JeffM]

1st line has -8y while the 2nd line has + 8y. To the OP, this was a silly error that we all make from time to time (it seems to happen to me way too often). However at the end the correct answer was obtain (ie -8y became 8y and then back to -8y)

Off to the corner for 8 minutes.

Thanks jomo.

 

[Steven G]

So that looks good, but technically it does not define a differentiable function.

I have a question to ask you. Suppose you have a non-function (ie two y for every x), but for any x the slope of the tangent line matches. Would you say that f'(x) exists?
To be very clear, suppose f had two rules. For any x, we get x^2 +3 AND x^2 + 11. Can we say that f'(x) = 2x????

EDIT: I guess it violates the definition of a derivative.

 

[Dr.Peterson]

So that looks good, but technically it does not define a differentiable function.
Why?

However, we are interested in a point above the x-axis, and the locus of points above the x-axis does correspond to the function
[MATH]y = \sqrt{\dfrac{4}{9} - x^2} + \dfrac{4}{3}.[/MATH]So what is dy/dx there?

I agree with Dr. Peterson that questions about dy/dt seem to come from outer space. But the questions seem to ask about them. Is there a part of the problem that you did not give us?

(1) I don't think the problem mentioned the path being a function at all, so that's not an issue. Nothing is said about dy/dx, either.
(2) The line of interest is on the y-axis; and your function does not include the point that was asked about, between the given points.
(3) I initially thought asking about dy/dt without mentioning a specific dependence on t was nonsense, but I erased everything I'd said when I realized dy/dt is in fact determined by the information provided.

 

[pka]

b) the point The distance the point is from \(\displaystyle \left(0,\dfrac{1}{3}\right)\) is where intersects the segment (0,0) to (0,1).

Rereading that I agree it makes no sense. I was being called to greet dinner guests.
I now think we all have misread this thread. I now think that the expected answer is a parametric equation \(\displaystyle (x(t),y(t))\)
\(\displaystyle \left(\frac{9}{4}\right)x^2+\left(\frac{9}{4}\right)\left(y-\frac{4}{3}\right)^2=1\) is the rectangular locus of the point.
Can you find the parametric form. If so then the \(\displaystyle \frac{dx}{dt}\) makes sense.

 

[Dr.Peterson]

I've made that kind of hasty post myself.

But although the problem is definitely about a parametric curve, we don't have any information to actually write a parametric equation (at least not in terms of time). The point is that, regardless of the specific functions x(t) and y(t), the locus constrains the derivatives with respect to t (given that they are never both zero, as stated). At the point where the curve crosses the y-axis, it is horizontal ...

 

[Miszka]

OK

[MATH]\sqrt{(x - 0)^2 + (y - 0)^2} = 2 * \sqrt{(x - 0)^2 + (y - 1)^2} \implies x^2 + y^2 = 4x^2 + 4y^2 - 8y + 4 \implies[/MATH]
[MATH]3y^2 + 8y + 3x^2 = -\ 4 \implies y^2 + 2 * \dfrac{4}{3} * y + x^2 = -\ \dfrac{4}{3} \implies[/MATH]
[MATH]\left \{ y^2 + 2\left ( \dfrac{4}{3} \right )y + \left ( \dfrac{4}{3} \right )^2 \right\} + x^2 = \dfrac{16}{9} - \dfrac{4}{3} = \dfrac{4}{9} \implies [/MATH]
[MATH]\left ( y - \dfrac{4}{3} \right )^2 + (x - 0)^2 = \left ( \dfrac{2}{3} \right )^2.[/MATH]
So that looks good, but technically it does not define a differentiable function.

Why?

However, we are interested in a point above the x-axis, and the locus of points above the x-axis does correspond to the function

[MATH]y = \sqrt{\dfrac{4}{9} - x^2} + \dfrac{4}{3}.[/MATH]
So what is dy/dx there?

I agree with Dr. Peterson that questions about dy/dt seem to come from outer space. But the questions seem to ask about them. Is there a part of the problem that you did not give us?

No, that's the whole exercise. I will try to solve it soon.