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posted by  anonymous2222 on 11/30/2008 2:15:36 AM  |  status: Live  

find dy/dx & the slope of the curve at indicated point

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Calculus N/A N/A N/A
Question Details:
find dy/dx & the slope of the curve at indicated point
 at (1,0)

my solution:








real answer:
2π

can you please explain what i did wrong and how to get 2π?

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posted by velixy on 11/30/2008 7:25:30 AM  |  status: Live
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Response Details:
find dy/dx & the slope of the curve at indicated point
 at (1,0)                                     REMEMBER THAT  (sin(u) )'= u'.cos(u)

y ' = 2 (πx-y)' . cos(πx-y)

y ' = 2 (π-y') . cos(πx-y)
 
 at (1,0)   y'(0) = 2(π -y' (0) )   cos(π-0)
 y'(0) = 2(π -y' (0) )   cos(π-0)                                          REMEMBER THAT  COS(π) = -1
 
y'(0) = - 2(π -y' (0) )  
 
y'(0) = - 2π +2y' (0) )  
 
-y'(0) = - 2π
y'(0) = 2π    answer
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posted by A-Rod 7821 on 11/30/2008 12:55:03 PM  |  status: Live
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anonymous2222's comment:
"thanks for the explanation"
Response Details:
 at the point (1, 0)
Use Implicit Differentiation
 
ANSWER:
 
Your work that you showed is all wrong. How did you get 0 on the left hand side? Using implicit differentiation, the derivative with respect to y is 1(dy/dx) not zero. Second of all you can not split up the right hand side like you did.  You have to use the chain rule.
So for . Do you understand this? The derivative of sine is cosine . Now take the derivative inside the parenthesis: The derivative of is  (taking the derivative with respect to x. The derivative with respect to -y is .
 
 
 
 
 
 
 
 
 
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