Find an equation of the tangent plane to the given parametric surface at the specified point.
r(u, v) = u cos vi + u sin vj + vk; u = 7, v = π/3
sqrt(3)/2(x-7/2)-1/2(y-7sqrt(3)/2)+7(z-pi/3)=0
To find the equation of the tangent plane to the given parametric surface at the specified point, we will follow these steps:
Step 1: Compute the partial derivatives of the position vector r(u, v).
Step 2: Evaluate the partial derivatives at the specified point.
Step 3: Use the partial derivatives to find the normal vector to the tangent plane.
Step 4: Use the normal vector and the specified point to write the equation of the tangent plane.
Now let's go through each step in detail:
Step 1: Compute the partial derivatives of the position vector r(u, v):
∂r/∂u = (cos v)i + (sin v)j + k
∂r/∂v = -u sin v i + u cos v j
Step 2: Evaluate the partial derivatives at the specified point:
∂r/∂u = (cos (π/3))i + (sin (π/3))j + k
= (1/2)i + (√3/2)j + k
∂r/∂v = -7 sin (π/3)i + 7 cos (π/3)j
= (-7/2)i + (7√3/2)j
Step 3: Use the partial derivatives to find the normal vector to the tangent plane.
The normal vector, N, is the cross product of the partial derivative vectors:
N = ∂r/∂u x ∂r/∂v
N = (1/2)i + (√3/2)j + k x (-7/2)i + (7√3/2)j
= 0i - (7/2)√3j + (√3/2)k
Step 4: Use the normal vector and the specified point to write the equation of the tangent plane.
The equation of the tangent plane is given by:
N · (r - r0) = 0
where N is the normal vector, r is the general position vector of a point on the plane, and r0 is the specified point.
At the specified point u = 7 and v = π/3,
r(u,v) = 7(cos(π/3)i + sin(π/3)j) + k
= (7/2)i + (7√3/2)j + k
The equation of the tangent plane becomes:
(0i - (7/2)√3j + (√3/2)k) · ((7/2)i + (7√3/2)j + k - (7/2)i - (7√3/2)j - k) = 0
Simplifying, we get:
(-7√3/2)j + ((√3/2)-(7√3/2))k = 0
The equation of the tangent plane is:
-7√3j - 3√3k = 0
To find the equation of the tangent plane to a parametric surface at a given point, we need to compute the partial derivatives of the surface with respect to u and v, and then evaluate them at the specified point.
Let's start by calculating the partial derivatives:
∂r/∂u = (cos v)i + (sin v)j + 0k
∂r/∂v = (-u sin v)i + (u cos v)j + 0k
Now, let's plug in the values u = 7 and v = π/3:
∂r/∂u = (cos (π/3))i + (sin (π/3))j + 0k
= (1/2)i + (√3/2)j + 0k
∂r/∂v = (-7 sin (π/3))i + (7 cos (π/3))j + 0k
= (-7√3/2)i + (7/2)j + 0k
So, the partial derivatives at the point (u, v) = (7, π/3) are:
∂r/∂u = (1/2)i + (√3/2)j + 0k
∂r/∂v = (-7√3/2)i + (7/2)j + 0k
Now, we can use these partial derivatives to find the equation of the tangent plane. The equation of a plane is given by:
Ax + By + Cz = D
where (A, B, C) is the normal vector to the plane, and D is a constant. The normal vector to the tangent plane is the cross product of the partial derivatives:
N = (∂r/∂u) × (∂r/∂v)
| i j k |
N = (1/2)i + (√3/2)j + 0k = | 1/2 √3/2 0 |
| -7√3/2 7/2 0 |
Now, we can substitute a point on the surface, (u, v) = (7, π/3), into the equation of the plane to find D.
r(7, π/3) = 7(cos(π/3))i + 7(sin(π/3))j + 1k
= (7/2)i + (7√3/2)j + 1k
Substituting these values into the equation of the plane:
(1/2)(7/2) + (√3/2)(7√3/2) + 0(D) = (7/2)(7/2) + (7√3/2)(7√3/2) + 0(1)
(49/4) + (63/4) + 0 = (49/4) + (63/4) + 0
112/4 = 112/4
28 = 28
Since both sides of the equation are equal, the equation of the tangent plane to the given parametric surface at the specified point (u, v) = (7, π/3) is:
(1/2)x + (√3/2)y + 0z = 28