Question

The plane y=1 intersects the surface z=x5+2xy−y5 in a certain curve. Find the slope of the tangent line of this curve at the point P=(1,1,2).

Answer 1

Answer:

slope = 7

Step-by-step explanation:

You have that the plane y=1 intersects the following surface:

$z=x^5+2xy-y^5$             (1)

To find the slope of the curve, you first replace y=1 in the equation (1):

$z=x^5+2x(1)-(1)^5=x^5+2x-1$       (2)

This is the generated curve when the plane y=1 intersect the surface z.

The slope of a function is given by the derivative of the function. Then, you calculate dz/dx in the equation (2):

$\frac{dz}{dx}=5x^4+2$

The slope only depends of the value of x. The slope for the point P(1,1,2) is:

$\frac{dz}{dx}_{x=1}=5(1)^4+2=7$

The value of the slope of the tangent line to point P is 7