Find an equation of the tangent plane (in the variables x, y and z) to the parametric surface
r(u,v) =(2u, 3u^2+5v, -4v^2) at the point (0,-10,-16)
To find an equation of the tangent plane to a parametric surface at a given point, we can use the following steps:
Step 1: Find the partial derivatives of the parametric surface with respect to u and v.
Step 2: Evaluate the partial derivatives at the given point.
Step 3: Use the evaluated partial derivatives and the given point to find the normal vector to the tangent plane.
Step 4: Write the equation of the tangent plane using the normal vector and the coordinates of the given point.
Let's go through each step in detail:
Step 1: Find the partial derivatives of the parametric surface with respect to u and v.
Given the parametric surface: r(u,v) = (2u, 3u^2 + 5v, -4v^2), we need to find the partial derivatives of this function. Taking the partial derivatives, we get:
∂/∂u (2u) = 2 and ∂/∂v (2u) = 0
∂/∂u (3u^2 + 5v) = 6u and ∂/∂v (3u^2 + 5v) = 5
∂/∂u (-4v^2) = 0 and ∂/∂v (-4v^2) = -8v
Step 2: Evaluate the partial derivatives at the given point.
The given point is (0, -10, -16), so we substitute these values of u, v into the partial derivatives we found in Step 1:
∂/∂u (2u) = 2(0) = 0 and ∂/∂v (2u) = 0(0) = 0
∂/∂u (3u^2 + 5v) = 6(0) = 0 and ∂/∂v (3u^2 + 5v) = 5
∂/∂u (-4v^2) = 0 and ∂/∂v (-4v^2) = -8(-16) = 128
So, the evaluated partial derivatives at the point (0, -10, -16) are:
∂/∂u = 0
∂/∂v = 5
∂^2/∂u∂v = 128
Step 3: Use the evaluated partial derivatives and the given point to find the normal vector to the tangent plane.
The normal vector to the tangent plane is given by the cross product of the two vectors formed by the partial derivatives. The normal vector is obtained using the formula:
n = (∂/∂u × ∂/∂v)
Calculating the cross product:
n = (0, 0, 1) × (1, 0, 128)
Using the cross product formula, we find:
n = (256, -128, 0)
Step 4: Write the equation of the tangent plane using the normal vector and the given point.
The equation of a plane can be written in the form ax + by + cz = d, where (x, y, z) are variables and a, b, c, d are constants. To write the equation of the tangent plane, we substitute the values of the given point and normal vector into the general equation:
256x - 128y + 0z = d
Substituting the coordinates of the given point (0, -10, -16), we get:
256(0) - 128(-10) + 0(-16) = d
1280 = d
Therefore, the equation of the tangent plane to the parametric surface at the point (0, -10, -16) is:
256x - 128y = 1280