# differentiation

• Aug 6th 2006, 10:53 PM
asian5
differentiation
differentiate
(X^2 + 200^2)^(1/2) multiplied by (1/4), and substitute X=200tanY

and

((500-X)^2 + 200^2)^(1/2) multiplied by (1/3), and substitute
X= -200tanZ+500

then combine equations and show that the minimum time occurs when (sinY)/m = (sinZ)/n where m and n are constants
• Aug 8th 2006, 10:06 AM
Jameson
Quote:

Originally Posted by asian5
differentiate
(X^2 + 200^2)^(1/2) multiplied by (1/4), and substitute X=200tanY

and

((500-X)^2 + 200^2)^(1/2) multiplied by (1/3), and substitute
X= -200tanZ+500

then combine equations and show that the minimum time occurs when (sinY)/m = (sinZ)/n where m and n are constants

So after making the substitution we have equations 1 and 2.

#1) $\frac{\sqrt{(200\tan(y)^2+200^2}}{4}$

#2) $\frac{\sqrt{200\tan(z)^2+200^2}}{3}$ (The 500's cancel out.)

Now a little bit of trig comes in.

#1) $\frac{1}{4} \times \sqrt{400\tan^2(y)+400}$
Factor out the 400, then take it out of the square-root function.

$\frac{20}{4} \times \sqrt{\tan^2(y)+1}$

Now make the trig substitution $\tan^2(x)+1=\sec^2(x)$

This should simplify things a lot and use the same process for equation #2.
• Aug 8th 2006, 08:49 PM
asian5
the other part of the question is tricky.. could you continue