Question

Let P be the tangent plane to the graph of (x,y)=20−10x2−20y2 at the point (4,2,−220). Let (x,y)=20−x2−y2. Find the point on the graph of where the tangent plane is parallel to P. (Use symbolic notation and fractions where needed. Give your answer in the form (*,*,*)).

Answer 1

To find the point on the graph where the tangent plane is parallel to P, we need to find the normal vector of the tangent plane and the point on the graph. The normal vector of the tangent plane is given by the coefficients of x, y, and z in the equation of the plane. The point on the graph where the tangent plane is parallel to P can be found by equating the equation of the plane and the equation of the graph.

Step-by-step explanation:

To find the point on the graph where the tangent plane is parallel to P, we need to find the normal vector of the tangent plane and the point on the graph. Let's start by finding the normal vector of the tangent plane P. The normal vector is given by the coefficients of x, y, and z in the equation of the plane. Since P is tangent to the graph at (4, 2, -220), the equation of P is of the form ax + by + cz = d, where a, b, and c are the coefficients of x, y, and z, respectively. Plugging in the values, we get:

4a + 2b - 220c = d

Next, let's find the equation of the graph. The equation is given by (x, y) = 20 - x^2 - y^2. To find the point on the graph where the tangent plane is parallel to P, the normal vector of the graph's tangent plane should be parallel to the normal vector of P. Since the normal vector of the graph's tangent plane is given by the coefficients of x, y, and z in the equation of the graph, we must have:

a = -2x

b = -2y

c = 1

Substituting these values into the equation of P, we get:

4(-2x) + 2(-2y) - 220(1) = d

Simplifying, we get:

-8x - 4y -220 = d

Now, we have the equations for both P and the graph. To find the point on the graph where the tangent plane is parallel to P, we can equate the two equations:

-8x - 4y - 220 = 4a + 2b - 220c

Substituting the values, we get:

-8x - 4y - 220 = 4(-2x) + 2(-2y) - 220(1)

Simplifying, we get:

-8x - 4y - 220 = -8x - 4y - 220

As we can see, the two equations are identical, meaning that the tangent plane is parallel to the graph at every point on the graph. Therefore, any point on the graph, such as (x, y) = (4, 2), would satisfy the condition that the tangent plane is parallel to P.

Answer 2

The x-coordinate of the point on the graph of f where the tangent plane is parallel to the tangent plane of g at (4, 2, -220) is x = 20.

To find the x-coordinate of the point on the graph of f where the tangent plane is parallel to the tangent plane of g at the point (4, 2, -220), we'll first find the equation of the tangent plane to g at that point.

The function g(x, y) = 20 - 10x² - 20y² represents a surface in three dimensions. The gradient of g at a point gives the normal vector to the tangent plane at that point.

The gradient of g is given by:

`\(\\abla g = \left((\partial g)/(\partial x), (\partial g)/(\partial y)\right)\)`

So, let's find the partial derivatives of g with respect to x and y:

`\((\partial g)/(\partial x) = -20x\)`

`\((\partial g)/(\partial y) = -40y\)`

At the point (4, 2):

`\((\partial g)/(\partial x) = -20(4) =-80`

`\((\partial g)/(\partial y) = -40(2) = -80\)`

Therefore, the normal vector to the tangent plane of g at (4, 2, -220) is
`\(\mathbf{n} = (-80, -80, 1)\)`.

Now, for a plane parallel to the tangent plane of g at (4, 2, -220), the normal vector will be the same. Let's find the gradient of f to determine where it matches the normal vector of the tangent plane of g.

The gradient of f is:

`\(\\abla f = \left((\partial f)/(\partial x), (\partial f)/(\partial y)\right)\)`

So, let's find the partial derivatives of f with respect to x and y:

`\((\partial f)/(\partial x) = -4x`

`\((\partial f)/(\partial y) = -2y\)`

For the normal vector of the tangent plane to f to be parallel to
`\(\mathbf{n} = (-80, -80, 1)\)`, we need:

`\((\partial f)/(\partial x) = -4x = -80\)`

Solving for x:

-4x = -80

x = 20