Question

Consider the following function. F(x,y,z)=2(x−7)² +(y−3)² +(z−6)² =10 Find the partial derivatives. Fx (x,y,z)=................ Fy (x,y,z)=.................. Fz (x,y,z)=................. Determine the partial derivatives at the point (8,5,8).

Fx (8,5,8)=
Fy​ (8,5,8)=
Fz (8,5,8)=
​
Now, consider the following. 2(x−7)² +(y−3)² +(z−6)² =10,(8,5,8) (a) Find an equation of the tangent plane to the given surface at the specified (b) Find an equation of the normal line to the given surface at the specified poi (x(t),y(t),z(t))=

Answer 1

The partial derivatives of the function F(x,y,z) = 2(x-7)² + (y-3)² + (z-6)² = 10 can be found by taking the derivative of each term. At the point (8,5,8), the partial derivatives are Fx(8,5,8) = 4, Fy(8,5,8) = 4, and Fz(8,5,8) = 4. The equation of the tangent plane at (8,5,8) is (x - 8)(Fx(8,5,8)) + (y - 5)(Fy(8,5,8)) + (z - 8)(Fz(8,5,8)) = 0, and the equation of the normal line is (x, y, z) = (8, 5, 8) + t(Fx(8,5,8), Fy(8,5,8), Fz(8,5,8)).

Step-by-step explanation:

To find the partial derivatives of the function F(x,y,z) = 2(x-7)² + (y-3)² + (z-6)² = 10, we need to take the derivative of each term with respect to the corresponding variable.

Fx(x,y,z) = 2 * 2(x-7) * 1 = 4(x-7)

Fy(x,y,z) = 2 * (y-3) * 1 = 2(y-3)

Fz(x,y,z) = 2 * (z-6) * 1 = 2(z-6)

At the point (8,5,8), we can substitute these values into the partial derivative expressions to find the values of the partial derivatives:

Fx(8,5,8) = 4(8-7) = 4

Fy(8,5,8) = 2(5-3) = 4

Fz(8,5,8) = 2(8-6) = 4

(a) To find an equation of the tangent plane to the given surface at the point (8,5,8), we can use the gradient of the function F(x,y,z) at that point. The equation of the tangent plane is given by:

(x - 8)(Fx(8,5,8)) + (y - 5)(Fy(8,5,8)) + (z - 8)(Fz(8,5,8)) = 0

(b) To find an equation of the normal line to the given surface at the point (8,5,8), we can use the normal vector of the tangent plane. The equation of the normal line is given by:

(x, y, z) = (8, 5, 8) + t(Fx(8,5,8), Fy(8,5,8), Fz(8,5,8))