Confirm Correctness of Ant's Shadow Movement Equation

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SUMMARY

The discussion centers on the calculation of the rate at which an ant's shadow moves on the y-axis, given its position and movement rates. The user initially calculated the shadow's movement rate as -9/16 but faced criticism for potential errors in their differentiation process. The correct approach involves using similar triangles to relate the shadow length and the ant's coordinates, leading to the conclusion that the correct rate is indeed -9/16, despite the user's confusion over signs in their calculations.

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Homework Statement



(Q) A lamp is placed at the point (5,0) and it casts the shadow of an ant onto the y axis. When the ant is at point (1,2), how fast is the ant's shadow moving when the ant's x-coordinate is increasing at the rate of 1/2 units/sec and its y-coordinate is decreasing at 1/5units/sec?

I got -9/16. Is it correct? I just need to confirm whether I'm right or wrong.


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The Attempt at a Solution

 
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That's not at all what I get. Too bad you didn't show your working- I might have been able to point out an error.
 
Here we go!

Let the coordinates of the ant be x,y. Let the length of the shadow be s.

Our goal is to find ds/dt. By similar triangles, s/y = 5/(5-x).

So, 5s - sx = 5y
Differentiating,

5(ds/dt) - (s(dx/dt) + x(ds/dt)) = 5(dy/dt)

When (x,y) = (1,2), s = 5/2.

5(ds/dt) - (5/2)(1/2) - (1)(ds/st) = 5(-1/5).

4(ds/dt) = -9/4.

ds/st = -9/16
 
mit_hacker said:
Let the coordinates of the ant be x,y. Let the length of the shadow be s.

Our goal is to find ds/dt. By similar triangles, s/y = 5/(5-x).

So, 5s - sx = 5y
Differentiating,

5(ds/dt) - (s(dx/dt) + x(ds/dt)) = 5(dy/dt)

When (x,y) = (1,2), s = 5/2.

5(ds/dt) - (5/2)(1/2) - (1)(ds/st) = 5(-1/5).

4(ds/dt) = -9/4.
You "lost a sign"
4(ds/dt)= +5/4- 1= 1/4, not -9/4.

ds/st = -9/16
 
Dammit!

All because of this gross gross gross error, I got the answer wrong in the exam. How many marks do you think I'll lose (out of 5?).??

Thanks a ton!
 

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