Find the point on the graph of z = 3x^2 − y^2 at which vector n = ⟨12, 4, −1⟩ is normal to the tangent plane.
So far I've got:
fx(x,y) = 6x = 6a
fy(x,y) = -2y = -2b
Where do I go from there?
To find the point on the graph of z = 3x^2 - y^2 where vector n = ⟨12, 4, -1⟩ is normal to the tangent plane, you need to find the values of x and y that satisfy both the partial derivative equations and the condition that the normal vector of the tangent plane is equal to vector n = ⟨12, 4, -1⟩.
Let's continue solving the partial derivatives:
fx(x, y) = 6x
fy(x, y) = -2y
Since we know that fx(x, y) = 6x and fy(x, y) = -2y, we can set them equal to the corresponding components of the normal vector n:
6x = 12 (equation 1)
-2y = 4 (equation 2)
Solve equation 1 for x:
6x = 12
x = 12/6
x = 2
Solve equation 2 for y:
-2y = 4
y = 4/(-2)
y = -2
Therefore, the point on the graph of z = 3x^2 - y^2 where vector n = ⟨12, 4, -1⟩ is normal to the tangent plane is (x, y, z) = (2, -2, z).
To find the value of z at this point, substitute the values of x and y into the equation for the graph:
z = 3(2^2) - (-2^2)
z = 3(4) - 4
z = 12 - 4
z = 8
Hence, the point on the graph of z = 3x^2 - y^2 where vector n = ⟨12, 4, -1⟩ is normal to the tangent plane is (2, -2, 8).
To find the point on the graph of z = 3x^2 − y^2 at which vector n = ⟨12, 4, −1⟩ is normal to the tangent plane, you need to find the gradient of the surface and then set it equal to the given normal vector.
First, find the partial derivatives of the surface equation:
fx(x, y) = 6x
fy(x, y) = -2y
Since you want to find the point where vector n is normal to the tangent plane, the gradient of the surface should be parallel to vector n. Therefore, the components of the gradient vector should be proportional to the components of n:
6x = 12 (1a)
-2y = 4 (1b)
0 = -1 (1c)
From equation (1a), solve for x:
6x = 12
x = 2
From equation (1b), solve for y:
-2y = 4
y = -2
Now substitute the values of x and y back into the equation of the surface to find z:
z = 3(2)^2 - (-2)^2
z = 12 - 4
z = 8
Therefore, the point on the graph where vector n is normal to the tangent plane is (2, -2, 8).