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April 23, 2014

April 23, 2014

Posted by **Lucy** on Tuesday, February 28, 2012 at 12:39am.

This is what Ive got so far...

dz/dt= 2x*dx/dt+2y*dy/dt

=4sin2tcos2t-2sintcost

I understand the steps of computing the max and min I just can't seem to solve this problem. Its on a test I have due tomorrow so please help!!

- Please help for my test CALculus -
**MathMate**, Tuesday, February 28, 2012 at 8:45amYou need to find dz/dt=0, so set

4sin2tcos2t-2sintcost =0

Expand using double-angle formulae:

8cos(t)sin(t)(cos(t)^2-sin(t)^2)-2cos(t)sin(t) = 0

Factorize:

-2cos(t)sin(t)(4sin(t)^2-4cos(t)^2+1)

Which gives

t=0, t=π/2, t=π and t=3π/2

when sin(t)=0 or cos(t)=0.

This leaves us with:

4sin(t)^2-4cos(t)^2+1=0

which can be reduced by the substitution

cos(t)^2=1-sin(t)^2

to a quadratic equation equivalent to

sin(t)=sqrt(3/8)

Now that leaves you to check each point as a maximum/minimum or inflection point using second derivatives.

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