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 need to find the normal vector to the surface at that point.
1. Find the partial derivatives of the position vector r(u,v) with respect to u and v:
∂r/∂u = (2, 6u, 0)
∂r/∂v = (0, 5, -8v)
2. Evaluate the partial derivatives at the point (0, -10, -16):
∂r/∂u(0, -10, -16) = (2, 0, 0)
∂r/∂v(0, -10, -16) = (0, 5, 80)
3. Take the cross product of the two partial derivatives to find the normal vector to the surface at the point:
n = (∂r/∂u(0, -10, -16)) × (∂r/∂v(0, -10, -16))
= (2, 0, 0) × (0, 5, 80)
= (-400, -160, 10)
4. Now we have the normal vector to the surface at the point (0, -10, -16), which is (-400, -160, 10). The equation of the tangent plane passing through this point is given by:
-400(x - 0) - 160(y + 10) + 10(z + 16) = 0
-400x - 160y + 10z + 1690 = 0
So, the equation of the tangent plane in the variables x, y, and z is -400x - 160y + 10z + 1690 = 0.