Approximation and error in functions of multiple variables - help needed

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SUMMARY

The discussion focuses on solving a problem involving the law of cosines and linear approximation to determine the distance a baseball fielder must move to catch a fly ball. The player is initially 310 feet from the camera, which turns 12 degrees to track the ball traveling 400 feet. The user sets up the function correctly but struggles with the linear approximation and calculating the maximum possible error due to positional inaccuracies of 5 feet and a 1-degree camera rotation error. The solution involves using the formula df = f_a da + f_b db + f_θ dθ to find the maximum error.

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Homework Statement


A baseball player is playing approximately 310 feet from a TV camera that is behind home plate. A batter hits a fly ball that goes to a wall 400 feet from the camera.
(a) The Camera turns 12 degrees to follow the play. Approximate the number of feet the fielder has to make the catch.
(b) If the position of the fielder could be in error by as much as 5 feet, and the maximum error in measuring the rotation of the camera is 1 degree, approximate the maximum possible error in your answer from part (a).

Verbatim.

Homework Equations


The law of cosines.
c2 = a2 + b2 - 2abcos(\theta)
Linear approximation at (a,b,c):
L(x,y,z) = f(a,b,c) + fx(x,y,z)(x-a) + fy(x,y,z)(y-b) + fz(x,y,z)(z-c)


The Attempt at a Solution



I used the law of cosines to set up a function of 3 variables, and I left one variable constant.

c = f(a,b,\theta) = the distance that the fielder moves.
a = the initial distance of the fielder from the camera.
b = the position of the ball after the play. This was constant at 400.
\theta = the angle the camera moves in radians.

f(a,b,\theta) = ( a2 + 4002 - 800*a*cos(\theta) )1/2

After that I used a linear approximation to approximate the distance the player must move (since the problem instructs me to do so). I got an answer nowhere near the exact value I calculated.

I do not know how to find the maximum possible error in (b).

I have the feeling that I did not set this problem up correctly to begin with. Did I?

EDIT: I got this feeling due to the fact that f(a,b,c) = 0 in my setup

fa = ( a - 400cos(\theta) ) / ( a2 +4002 -a*800cos(\theta) )1/2
f\theta = ( 400asin(\theta) ) / ( a2 +4002 -a*800cos(\theta) )1/2
 
Last edited:
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I would NOT interpret "approximate the number of feet" as meaning to use a linear approximation. You will be approximating by simply calculating f(a,b,\theta) since the value of cos(12) that your calculator gives you is only approximate.

For (b) you can use the approximation df= f_a da+ f_b db+ f_\theta d\theta, with da= db= 5 and d\theta= 1 degree (so you will need to convert this to radians).
 
Thank you. That clears things up.
 

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