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

Answer:

Options a and d are the correct ones.

Step-by-step 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

$z=f(x_0,y_0)+\frac{df}{dx}(x_0,y_0) (x-x_0)+\frac{df}{dy}(x_0,y_0) (y-y_0)$

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

$\frac{df}{dx} = 2x, \frac{df}{dy} = -3y^2$

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

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