Calculus and Derivatives

Question

Given the function y = 11 sin (7 pi x - 5 y) and the point (4,0). (A) Verify that the given point is on curve. (B) Find the equation of the tangent line to the curve at the given point. (C) Find the equation of the normal line to the curve at the given point.

Answer 1

y = 11sin(7pix -5y) verify (4,0) is on the curve

0 = 11sin(28pi) = 11sin2pi = 11(0) = 0. it's on the sine curve

tangent line has slope = the derivative of y = y' = dy/dx, at the point x=4, y=0

y'(x) = 11cos(7pix)

y'(4) = 11cos(28pi) = 11=m

tangent line is y=mx +b = y=11x +b to find b, plug in the point (4,0)

0 =44+b

b =-44

tangent line is y = 11x -44x

normal line has negative inverse slope = -1/11

y = -x/11 + b, plug in (4,0) to find b

0 = -4/11 +b, b = 4/11

normal line is y = -x/11 +4/11 in slope intercept form

or

x +11y = 4 in standard form ax+by=c where a=1, b=11, c = 4