Related Rates Double Check?

[eric_f]

Hi All,

I've completed this problem, but am second guessing myself on the answer. Can someone explain why my answer is not the same as if the light were running around a circular wall with radius same as the hypotenuse of the triangle in the solution? According to my answer, the light is travelling at roughly 8,168 meters/min, but if it were on a fixed circle with radius of sqrt(130,000) meters, it would only travel at 6,796 meters/min. Shouldn't the instantaneous velocities be equal?



Thanks for the insight!

~Eric

 

[Deleted member 4993]

Hi All,

I've completed this problem, but am second guessing myself on the answer. Can someone explain why my answer is not the same as if the light were running around a circular wall with radius same as the hypotenuse of the triangle in the solution? According to my answer, the light is travelling at roughly 8,168 meters/min, but if it were on a fixed circle with radius of sqrt(130,000) meters, it would only travel at 6,796 meters/min. Shouldn't the instantaneous velocities be equal?

View attachment 2959

Thanks for the insight!

~Eric

Your image too light - very hard to read!!

 

[HallsofIvy]

You have done the problem correctly. As for "why my answer is not the same as if the light were running around a circular wall with radius same as the hypotenuse of the triangle in the solution?", imagine a white wall, 20 feet wide, say. about 10 feet from you, with a glass wall 5 feet in front of it. You are 10 feet from the exact center of the white wall which extends 10 feet on either side of you.

Shine a flash light at the left corner of the white wall. Do you see that the light beam will will pass through the glass wall closer than 10 feet to your left? You can, in fact, use geometry- similar triangles (which is the basis for trigonometry)- to find that distance. The white wall gives a right triangle with legs of length 10 ft each. The triangle formed by the glass wall is 5 feet from you so, that is a right triangle with one leg (the distance from you) of length 5 ft and so, by "similar triangles", the other, the distance to your left of 5 ft also. And, of course, if you shine the flash light at the right edge of the white wall, 10 feet to your right, the light will go through the glass at a point 5 feet to your right.

That is, in the time that you have swung the light from left to right a total of 10+ 10= 20 feet, the point where the light passes through the glass have moved 5+ 5= 10 feet. As you swing the light through an arc, the speed at which the "spot of light" on a wall moves, depends upon the distance from the source of light.

If you had a circular wall, centered at the source of light, the speed at which the spot of light moves is constant because the distance is contstant. If you have a straight wall, as in this problem, the distance from the source of light to the wall varies so the speed varies.

(In fact, when you say "with radius same as the hypotenuse of the triangle in the solution", do you not see that there is not ONE triangle, but a different right triangle for each angle?)