For the function f (x,y) = y2 - 2x + 4 sin (3x + 2y) , find the vector equation

Question

For the function f (x,y) = y2 - 2x + 4 sin (3x + 2y) , find the vector equation of the line tangent to the curve z = f (2, y) at the point (2,1, f (2,1)) (this line is in the plane x = 2) (Continuation of #1). Use Mathematica to plot the surface z = f(x, y), the curve z = f (2, y) and the tangent line you found in Problem 1. Use a PlotRange so that 0lexle3, 0leyle2, -10 le 2 le 10, and use the option BoxRatios -> {1, 1, 1}- Rotate so the tangency is clear. Print this plot and your code. The Lab Two code will be helpful!

Explanation / Answer

       To find the vector equation of the line tangent to the curve

          z = f(2,y) at the point (2,,1,f(2,1))

           We have, f(x,y) = y^2 -2x + 4sin(3x+2y)

               f(x,y,z) = y^2 -2x + 4sin(3x+2y) -z

                           Grad[f(x,y,z)] = [ -2+12cos(3x+2y) , 2y +8cos(3x+2y) , -1]

Grad[f(x,y,z)]    at [2,1, -3+4sin8]    is

                        = [ -2+12cos8,2+8cos8 , -1]

              Now, vector equation of the line is

                           r(t) = [ -2+12cos8,2+8cos8 , -1] + t [ 2,1,-3+4sin8]

                          

              

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