Explanation:
To find the equation of the tangent plane to the surface at the point (5 cos(π/3), 5 sin(π/3), k), we need to first find the partial derivatives of r with respect to u and v:
ru(u, v) = cos(v) i + sin(v) j
rv(u, v) = -5 sin(v) i + 5 cos(v) j + k k
Then, we can plug in the point (5 cos(π/3), 5 sin(π/3), k) to get the partial derivatives evaluated at that point:
ru(5, π/3) = cos(π/3) i + sin(π/3) j = (1/2) i + (√3/2) j
rv(5, π/3) = -5 sin(π/3) i + 5 cos(π/3) j + k k = -5/2 i + (5√3/2) j + k k
Now we can use the point-normal form of the equation of a plane, where the normal vector is given by the cross product of ru and rv:
⟨ru(5, π/3) × rv(5, π/3)⟩ = ⟨(1/2) i + (√3/2) j, (-5/2) i + (5√3/2) j, k⟩ = (-5/2 - √3/2) i + (-5√3/2 - 1/2) j + k k
Thus the equation of the tangent plane to the surface at (5 cos(π/3), 5 sin(π/3), k) is:
(-5/2 - √3/2)(x - 5 cos(π/3)) + (-5√3/2 - 1/2)(y - 5 sin(π/3)) + (z - k) = 0
Simplifying, we get:
-5√3 x + 5y - 5√3 z + (25/2 + 5/2√3) = 0
So the equation of the tangent plane to the surface at the point (5 cos(π/3), 5 sin(π/3), k) is:
-5√3 x + 5y - 5√3 z + 15.79 = 0.