Asked by Anonymous
Use implicit differentiation to show that a function defined implicitly by sin x +
cos y = 2y has a critical point whenever cos x = 0. Then use the first derivative
test to classify those critical numbers that lies in the interval (−2, 2) as relative
maxima or minima.
I KNOW HOW TO DO THE IMPLICIT PART BUT
PLIZ CAN ANY 1 EXPLAIN HOW TO FIND THE CRITTICAL POINTS FOR ME..
cos y = 2y has a critical point whenever cos x = 0. Then use the first derivative
test to classify those critical numbers that lies in the interval (−2, 2) as relative
maxima or minima.
I KNOW HOW TO DO THE IMPLICIT PART BUT
PLIZ CAN ANY 1 EXPLAIN HOW TO FIND THE CRITTICAL POINTS FOR ME..
Answers
Answered by
Ms. Sue
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Answered by
Reiny
if sinx + cosy = 2y
then cosx - siny dy/dx = 2dy/dx
cosx = 2dy/dx + siny dy/dx
dy/dx(2 - siny)= cosx
dy/dx = cosx/(2-siny)
for critical values, dy/dx = 0
thus cosx =0 , as required.
if cosx = 0 , then x = π/2 or x = 3π/2 or
if x=π/2 , into original ....
sin π/2 + cosy = 2y
1 + cosy = 2y
cosy = 2y - 1 , messy equation to solve (y = appr .83543 )
but it did not say to find the actual point but rather to classify as max/min
so second derivative:
y'' = ( (2-siny)(-sinx) - cosx(-cosy dy/dx)/ (2-siny)^2
if x = π/2
y'' = ( 2 - (# less than 1)(-1) - 0(....) / (positive)
= -/+ = negative
so when x = π/2 , we have a maximum
if x = 3π/2
y'' = (2 - .....)((+1) - 0(....) )/(positive)
= +/+ = +
so when x = 3π/2, we have a minimum
Just realized we are to looks at interval (-2,2)
so for cosx = 0
x = -π/2, π/2
Repeating an evaluation similar to the ones above,
when x = -π/2
y'' = +/+ = +
we would have a minimum
then cosx - siny dy/dx = 2dy/dx
cosx = 2dy/dx + siny dy/dx
dy/dx(2 - siny)= cosx
dy/dx = cosx/(2-siny)
for critical values, dy/dx = 0
thus cosx =0 , as required.
if cosx = 0 , then x = π/2 or x = 3π/2 or
if x=π/2 , into original ....
sin π/2 + cosy = 2y
1 + cosy = 2y
cosy = 2y - 1 , messy equation to solve (y = appr .83543 )
but it did not say to find the actual point but rather to classify as max/min
so second derivative:
y'' = ( (2-siny)(-sinx) - cosx(-cosy dy/dx)/ (2-siny)^2
if x = π/2
y'' = ( 2 - (# less than 1)(-1) - 0(....) / (positive)
= -/+ = negative
so when x = π/2 , we have a maximum
if x = 3π/2
y'' = (2 - .....)((+1) - 0(....) )/(positive)
= +/+ = +
so when x = 3π/2, we have a minimum
Just realized we are to looks at interval (-2,2)
so for cosx = 0
x = -π/2, π/2
Repeating an evaluation similar to the ones above,
when x = -π/2
y'' = +/+ = +
we would have a minimum
Answered by
Jay
critical point are when the derivative is equal to 0
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