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A differentiable function f(x,y)f(x,y) has the property that f(2,5)=5f(2,5)=5 and fx(2,5)=−7fx(2,5)=−7 and fy(2,5)=7fy(2,5)=7. Find the equation of the tangent plane at the point on the surface z=f(x,y)z=f(x,y) where x=2x=2, y=5y=5.

Respuesta :

sqdancefan

Answer:

  -7x +7y -z = 16

Step-by-step explanation:

We can define the function ...

  F(x, y, z) = f(x, y) -z

and differentiate at the point (x, y, z) = (2, 5, 5) to get ...

  fx(2, 5, 5) = -7 . . . . given

  fy(2, 5, 5) = 7 . . . . given

  fz(2, 5, 5) = -1 . . . . partial derivative of the above equation

Then the equation of the plane can be written as ...

  fx(x -2) +fy(y -5) +fz(z -5) = 0

  -7(x -2) +7(y -5) -1(z -5) = 0 . . . . . substitute for fx, fy, fz

  -7x +14 +7y -35 -z +5 = 0 . . . . . eliminate parentheses

  -7x +7y -z = 16 . . . . equation of the tangent plane

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