Mathematics, 08.04.2020 00:12 jesussaves333

(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.

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Answer from: khanhlan7213

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)

Answer from: Quest

1/4 because there is 4 grades.

Answer from: Quest

answer: hey there,

the answer you are looking for is a..

hope this you!

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(1 point) A student was asked to find the equation of the tangent plane to the surface z=x2−y3 at th...