find an equation of the tangent line to the curve y=x√x that is parallel to the line y=3+6x

Question

Need help to sold the Calculus problem  find an equation of the tangent line to the curve y=x√x that is parallel to the line y=3+6x

Answer 1

Slope of the line y = 6x+3 is 6

 

y = x√x = x^3/2

 

   y' = (3/2)x^1/2 = slope of tangent line at (x, x^3/2)

 

   Since the tangent line is parallel to the line y = 6x + 3, (3/2)x^1/2 = 6

 

                                                                                          x^1/2 = 4

 

                                                                                              x = 16

 

   Point on tangent line:  (16, 16^3/2) = (16, 64)

 

   So, the tangent line has slope 6 and passes through the point (16,64).

 

       Equation of tangent line:  y - 64 = 6(x - 16)

 

                                                    y = 6x - 32