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..
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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
critical point are when the derivative is equal to 0
To find the critical points of a function, we need to solve for the values of x and y where the derivative is either zero or undefined. In this case, we are given that cos x = 0, so let's start by finding the values of x that satisfy this equation.
Since cos x = 0 when x = π/2 + nπ, where n is an integer, we can substitute these values of x back into the original equation sin x + cos y = 2y to find the corresponding values of y.
For x = π/2 + nπ, we have sin(π/2 + nπ) + cos y = 2y. Let's solve for y in terms of n.
sin(π/2 + nπ) + cos y = 2y
sin(π/2) + cos y = 2y (since sin(nπ) = 0 for all integers n)
1 + cos y = 2y
cos y = 2y - 1
Now, in order to find the critical points, we need to solve for y in terms of n.
cos y = 2y - 1
y = (cos y + 1) / 2
So, for each value of x = π/2 + nπ, the corresponding y value is given by y = (cos y + 1) / 2.
To classify the critical numbers as relative maxima or minima, we can apply the first derivative test. This involves taking the derivative of the function and analyzing the sign of the derivative at the critical points.
Differentiating the original equation sin x + cos y = 2y implicitly, we get:
cos x + (-sin y * dy/dx) = 2(dy/dx)
Rearranging the equation, we have:
dy/dx = (cos x - 2) / (-sin y)
Now, substitute the critical points (x = π/2 + nπ, y = (cos y + 1) / 2) into the dy/dx equation and analyze the sign of the resulting derivative.
dy/dx = (cos (π/2 + nπ) - 2) / (-sin ((cos (π/2 + nπ) + 1) / 2))
For each value of n, calculate this derivative and determine the sign. If the derivative is positive (greater than zero), then the corresponding critical point is a relative minimum. If the derivative is negative (less than zero), then the corresponding critical point is a relative maximum.
Evaluate dy/dx for the critical numbers n = -2, -1, 0, 1, and 2, and based on the sign of each derivative, determine whether each critical number is a relative maximum or minimum in the given interval (-2π, 2π).