MATH SOLVE

4 months ago

Q:
# A student was asked to find the equation of the tangent plane to the surface z=x3−y4 at the point (x,y)=(1,3). The student's answer was z=−80+3x2(x−1)−(4y3)(y−3). (a) At a glance, how do you know this is wrong. What mistakes did the student make? Select all that apply. A. The (x - 1) and (y - 3) should be x and y. B. The -80 should not be in the answer. C. The partial derivatives were not evaluated a the point. D. The answer is not a linear function. E. All of the above (b) Find the correct equation for the tangent plane. z=

Accepted Solution

A:

Answer:C) The partial derivatives were not evaluated a the point.D) The answer is not a linear function.The correct equation for the tangent plane is [tex]z = 241 + 3 x - 108 y[/tex] or [tex]3x-108y-z+241 = 0[/tex]Step-by-step explanation:The equation of the tangent plane to a surface given by the function [tex]S=f(x, y)[/tex] in a given point [tex](x_0, y_0, z_0)[/tex] can be obtained using:[tex]z-z_0=f_x(x_0,y_0,z_0)(x-x_0)+f_y(x_0,y_0, z_0)(y-y_0)[/tex] (1)where [tex]f_x(x_0, y_0, z_0)[/tex] and [tex]f_y(x_0, y_0, z_0)[/tex] are the partial derivatives of [tex]f(x,y)[/tex] with respect to [tex]x[/tex] and [tex]y[/tex] respectively and evaluated at the point [tex](x_0, y_0, z_0)[/tex].Therefore we need to find two missing inputs in our problem in order to use equation (1). The [tex]z_0[/tex] coordinate and the partial derivatives [tex]f_x(x_0, y_0, z_0)[/tex] and [tex]f_y(x_0, y_0, z_0)[/tex]. For [tex]z_0[/tex] just evaluating in the given function we obtain [tex]z_0= -80[/tex] and the partial derivatives are:[tex]\frac{\partial f(x,y)}{\partial x} \equiv f_x(x, y)= 3x^2 \\f_x(x_0, y_0) = f_x(1, 3) = 3[/tex][tex]\frac{\partial f(x,y)}{\partial y} \equiv f_y(x, y)= -4y^3 \\f_y(x_0, y_0) = f_y(1, 3) = -108[/tex]Now, substituting in (1)[tex]z-z_0=f_x(x_0,y_0,z_0)(x-x_0)+f_y(x_0,y_0, z_0)(y-y_0)\\z + 80 = 3x^2(x-1) - 4y^3(y-3)\\z = -80 + 3x^2(x-1) - 4y^3(y-3)[/tex]Notice that until this point, we obtain the same equation as the student, however, we have not evaluated the partial derivatives and therefore this is not the equation of the plane and this is not a linear function because it contains the terms ([tex]x^3[/tex] and [tex]y^4[/tex])For finding the right equation of the tangent plane, let's substitute the values of the partial derivatives evaluated at the given point:[tex]z = -80 + 3x^2(x-1) - 4y^3(y-3)\\z = -80 +3(x-1)-108(y-3)\\z = 241 + 3 x - 108 y[/tex]or [tex]3x-108y-z+241 = 0[/tex]