Question

Find the equation of the tangent plane to the graph of f(x,y)=ln(11x

2
−8y
2
) at the point (1,1). (Use symbolic notation and fractions where needed. Enter your answer using x−,y-, z-coordinates.)

Answer 1

To find the equation of the tangent plane to the graph of f(x,y)=ln(11x^2-8y^2) at the point (1,1), we need to find the partial derivatives of f(x,y) and substitute the given point into these derivatives. Using the partial derivatives, we can determine the equation of the tangent plane as z-(ln(3/100))=(22/3)(x-1)-(16/3)(y-1).

Step-by-step explanation:

To find the equation of the tangent plane to the graph of f(x,y)=ln(11x2-8y2) at the point (1,1), we need to find the values of fx, fy, and f(1,1). Taking the partial derivatives, we get fx=22/(11x2-8y2) and fy=(-16y)/(11x2-8y2). Substituting (1,1) into these equations, we get fx=22/3 and fy=-16/3.

So, the equation of the tangent plane is given by (z-f(1,1))=fx(x-1)+fy(y-1), which simplifies to z-(ln(3/100))=(22/3)(x-1)-(16/3)(y-1).Using the partial derivatives, we can determine the equation of the tangent plane as z-(ln(3/100))=(22/3)(x-1)-(16/3)(y-1).

Answer 2

The equation of the tangent plane to the graph of
`\(f(x, y) = \ln(11x^2 - 8y^2)\)` at the point (1, 1) is
`\(z - \ln(3) = (22)/(3) (x - 1) - (16)/(3) (y - 1)\)`.

To find the equation of the tangent plane to the graph of
`\(f(x, y) = \ln(11x^2 - 8y^2)\)` at the point (1, 1), we'll first find the partial derivatives of f with respect to x and y, and then use them to construct the equation of the tangent plane.

Given function:
`\(f(x, y) = \ln(11x^2 - 8y^2)\)`

Partial derivatives:

`\((\partial f)/(\partial x)\)` (partial derivative of f with respect to x) and
`\((\partial f)/(\partial y)\)` (partial derivative of f with respect to y).

`\((\partial f)/(\partial x) = (22x)/(11x^2 - 8y^2)\)`

`\((\partial f)/(\partial y) = (-16y)/(11x^2 - 8y^2)\)`

Now, to find the equation of the tangent plane at the point (1, 1), we'll calculate the values of the partial derivatives at this point.

At (1, 1):

`\((\partial f)/(\partial x)\)` evaluated at (1, 1):

`\((22 * 1)/(11 * 1^2 - 8 * 1^2) = (22)/(3)\)`

`\((\partial f)/(\partial y)\)` evaluated at (1, 1):

`\((-16 * 1)/(11 * 1^2 - 8 * 1^2) = -(16)/(3)\)`

The equation of the tangent plane is given by:

`\[z - f(1, 1) = (\partial f)/(\partial x)(1, 1) \cdot (x - 1) + (\partial f)/(\partial y)(1, 1) \cdot (y - 1)\]`

Now, substitute the values into the equation:

`\[z - \ln(11 * 1^2 - 8 * 1^2) = (22)/(3) \cdot (x - 1) - (16)/(3) \cdot (y - 1)\]`

Simplify further:

`\[z - \ln(3) = (22)/(3) (x - 1) - (16)/(3) (y - 1)\]`

Question:

Find the equation of the tangent plane to the graph of
`f(x,y) = ln(11x^2-8y^2)` at the point (1,1). (Use symbolic notation and fractions where needed.) z = ____