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August 19th, 2016, 08:35 PM   #1
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Instantaneous Rate of Change Problem

Okay so I am trying to figure this problem here.

On the attachment photo I have below, imagine that point C begins to move to the right at a rate of 1. At what rate does point d move, right when point C begins to move.

Obviously as point C continues at a rate of 1, the rate of point d changes. But my question is, what is the rate of point d right when point C starts moving? What is the instantaneous rate of change immediately when point C starts moving to the right at a rate of 1?

I'm sure there are some clever enough minds to figure this one out.
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August 20th, 2016, 06:07 AM   #2
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How far along AB is d? That is, what is the value of Ad/AB?
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August 20th, 2016, 10:34 AM   #3
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Shouldn't it be 0? C is moving by the rate of 1 so dC=t
When t = 1 it has moved 1 unit. For t = 0 it hasn't moved yet, therefor point D hasn't moved either
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August 20th, 2016, 10:38 AM   #4
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Oh wait, this is high school math, sorry. Explanation of previous post: point C is moving in time by 1 unit in 1 unit of time. So: movement of C = unit of time. For time = 0 (when it started to move) it is equal to 0, it hasn't moved yet in that instant.

Last edited by skipjack; August 20th, 2016 at 05:08 PM.
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August 20th, 2016, 12:03 PM   #5
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Sorry for the stupid things I wrote. They are WRONG. This should be it. I only wrote the x coordinate of the point d. If I had numbers it would look a bit nicer. Sorry for the handwriting.
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Last edited by skipjack; August 20th, 2016 at 05:08 PM.
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August 20th, 2016, 12:18 PM   #6
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Explanation of what I wrote. First I found the equation of the line AB using the formula. Then I found the equation of the lime passing through origin and the point
c(x3+t, y3+t) (+t for the rate of change). We want to find how coordinates of point d are changing, so we need to solve the system of equations to find its points. Also, I've made a mistake when I applied the minus to the fraction.

Last edited by skipjack; August 20th, 2016 at 05:09 PM.
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August 20th, 2016, 05:13 PM   #7
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Quote:
Originally Posted by VisionaryLen View Post
I am trying to figure this problem here.
Are A and B fixed points?
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August 20th, 2016, 06:34 PM   #8
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1) The diagram is ambiguous, it suggests some things that aren't stated. May we get clarification from the OP?

* Is 0dB a right angle?

* May we assume that if 0dB is a right angle, then given B, A and C must be chosen on some horizontal line (whose height is yet another variable, I think) such that A and B are symmetric with respect to the top corners of the square made by raising a perpendicular from B?

* Or is this not a square at all?

2) I think this is just a related rate problem. I was always terrible at those.

3) I throw this out in case it helps anyone. Say you flip the diagram vertically then horizontally like this:



Now ignore those '1''s on those little triangles, and imagine the unit circle centered at the origin with radius 1, which we imagine is the position of the vertical line AC.

Then the height of C is the tangent of the angle C-origin-(1,0). And if the distance from the origin to the crossing point of AC with the x-axis is k, then the height of C is k tan theta.

So a lot depends I think on whether the central crossing is a right angle or not. Because given C, you then choose A and draw the perpendicular to 0-C and extend to B.
Perhaps something here is helpful.

Last edited by Maschke; August 20th, 2016 at 06:39 PM.
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