Answer to Question #340924 in Calculus for ser

Question #340924

: Let ๐‘“(๐‘ฅ, ๐‘ฆ) = ๐‘ฅ^2 + ๐‘ฆ^3 . Find the slope of the line tangent to this surface at the point (-1, 1) and lying in the plane x = -1


1
Expert's answer
2022-05-17T00:20:16-0400

"f\\left(x,y\\right)=x^2+y^3" , "M=\\left(-1,1\\right)" ,

"df\\left(x,y\\right)=\\frac{\\partial f}{\\partial x}dx+\\frac{\\partial f}{\\partial y}dy=2xdx+{3y}^2dy" ,

The equation of a plane tangent to a point "M=\\left(x_0,y_0\\right)" of the surfaceย  "z=f\\left(x,y\\right)" ย has the form:

"z-z_0={f\\prime}_x\\left(x_0,y_0\\right)\\left(x-x_0\\right)+{f\\prime}_y\\left(x_0,y_0\\right)\\left(y-y_0\\right),\\ z_0=f\\left(x_0,y_0\\right)" ,

"z_0=f\\left(x_0,y_0\\right)=2, {f\\prime}_x\\left(x_0,y_0\\right)=-2, {f\\prime}_y\\left(x_0,y_0\\right)=3" ,

Then

"z-2=-2\\left(x+1\\right)+3\\left(y-1\\right)" ,

By "x=-1" given:

"z-2=3\\left(y-1\\right)=k\\left(y-1\\right),\\ k=3".

Answer: the slope of the line tangent to this surface at the point (-1, 1) and lying in the plane x=โ€“1 is k=3


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS