How do you find the slope of the line tangent to #2sin^2x=3cosy# at #(pi/3,pi/3)#?

Question

Samantha Adkison · Accepted Answer

#(2/3)cos(x)/sin(cos^-1((2/3)sin(x)))=dy/dx#

Divide by 3 on both sides:

#(2/3)sin(x)=cos(y)#

Then use implicit differentiation. Note that the derivative of #sin(x)# is #cos(x)# and the derivative of #cos(x)# is #-sin(x)#. The #(2/3)# is unaffected

#(2/3)cos(x)=-sin(y) (dy/dx)#

Solve for #dy/dx#

#(2/3)cos(x)/sin(y)=dy/dx#

However, you can solve for y from the given equation and plug it in for y.

#(2/3)sin(x)=cos(y)#

#cos^-1((2/3)sin(x))=y#

#(2/3)cos(x)/sin(cos^-1((2/3)sin(x)))=dy/dx#

Ezra Adams · Answer

undefined To find the slope of the line tangent to the equation 2sin^2x = 3cosy at the point (pi/3, pi/3), we need to differentiate both sides of the equation with respect to x and y, respectively.

Differentiating 2sin^2x with respect to x gives us 4sinxcosx.

Differentiating 3cosy with respect to y gives us -3siny.

Next, we substitute the given point (pi/3, pi/3) into the derivatives we obtained.

For the derivative of 2sin^2x, we have sin(pi/3) = sqrt(3)/2 and cos(pi/3) = 1/2.

For the derivative of 3cosy, we have cos(pi/3) = 1/2.

Substituting these values, we get 4(sqrt(3)/2)(1/2) = 2sqrt(3) for the derivative of 2sin^2x, and -3(1/2) = -3/2 for the derivative of 3cosy.

Therefore, the slope of the line tangent to the equation 2sin^2x = 3cosy at the point (pi/3, pi/3) is 2sqrt(3)/(-3/2), which simplifies to -4sqrt(3)/3.