Question

(1 point) A student was asked to find the equation of the tangent plane to the surface z=x2−y3 at the point (x,y)=(5,1). The student's answer was z=24+2x(x−5)−(3y2)(y−1). (a) At a glance, how do you know this is wrong. What mistakes did the student make? Select all that apply.

Answer 1

Options a and d are the correct ones.

Explanation:

The complete question is as follows

A student was asked to find the equation of the tangent plane to the surface
`z = x^2-y^3` at the point (x,y)=(5,1). The student's answer was
`z = 24+2x(x-5)-3y^2(y-1)`. (a) At a glance, how do you know this is wrong. What mistakes did the student make? Select all that apply.

a) The answer is not a linear function

b) The (x-5) and (y-1) should be x and y

c) the 24 should not be in the answer

d) the partial derivatives were not evaluated at the point

e) all of the above.

Recall that, given a surface of the form
`z=f(x,y)` where f is differentiable, then at a given point
`(x_0,y_0)` we can find the equation of the tangent plane by

`z=f(x_0,y_0)+(df)/(dx)(x_0,y_0) (x-x_0)+(df)/(dy)(x_0,y_0) (y-y_0)`

We are given that
`(x_0,y_0) = (5,1)` Note that f(5,1) = 24. Then, c is not true. Also, the formula says that we must have the factors (x-5) and (y-1), since we are evaluating the tangent plane in a neighborhood of that point, hence option b is not true. Since B and C are not true, then E is not true. Note that the given function by the student has a mutiplication of 2x and (x-5) which will give us the function
`2x^2-10` which is of degree two. Then, the function given by the student is not a linear function, since linear functions have a degree at most 1. Finally, we must check that d is also true.

REcall that

`(df)/(dx) = 2x, (df)/(dy) = -3y^2`

so we see that this coincide with the the terms that multiply the factors (x-5) and (y-1) respectively, which tell us that even though the student was trying to follow the formula, he/she forgot to evaluate the partial derivatives at the given point, hence the option D is also true.

Recall that if we want the tangent plane at the point given, we must evaluate the partial derivatives at x=5 and y=1. Hence, the formula of the tangent plane is

`z = 24+10(x-5)-3(y-1)`