Use implicit differentiation to find an equation of the tangent line to the curve at the given point. ?
1 Answer
Mar 19, 2018
Explanation:
We have using implicit differentiation:
#1(sin(16x))(dy/dx) + y(16cos(16x)) = 1(cos(2y)) - 2xsin(2y)(dy/dx)#
#sin(16x)(dy/dx) + 2xsin(2y)(dy/dx) = cos(2y) - 16ycos(16x)#
#dy/dx = (cos(2y) - 16ycos(16x))/(sin(16x) + 2xsin(2y))#
We now find the slope of the tangent at that point.
#dy/dx|_((pi/2, pi/4)) = (cos(2(pi/4)) - 16(pi/4)cos(16(pi/2)))/(sin(16(pi/2)) + 2(pi/2)sin(2(pi/4))#
#dy/dx|_((pi/2, pi/4)) = (0 - 4pi)/(0 + pi)#
#dy/dx|_((pi/2, pi/4)) = -4#
Now for the equation
#y - pi/4 = -4(x- pi/2)#
#y = -4x + 2pi + pi/4 = -4x + (9pi)/4#
Hopefully this helps!