Question

Find an equation of the tangent plane to the given parametric surface at the specified point.

r(u, v) = u2 i + 9u sin v j + u cos v k; u = 2, v = 0

Answer 1

The equation of the tangent plane to the given parametric surface
`\(r(u, v) = u^2 \mathbf{i} + 9u \sin v \mathbf{j} + u \cos v \mathbf{k}\)`at the specified point
`\(u = 2, v = 0\) is \(x - 4y - 3z + 11 = 0\).`

Step-by-step explanation:

The equation of the tangent plane to a parametric surface at a specific point is given by the dot product of the normal vector and the vector from the point to the parametric surface.

The parametric surface is defined as
`\(r(u, v) = u^2 \mathbf{i} + 9u \sin v \mathbf{j} + u \cos v \mathbf{k}\)`. To find the tangent plane at the point \(u = 2, v = 0\), we first calculate the partial derivatives of \(r(u, v)\) with respect to \(u\) and \(v\). The cross product of these partial derivatives gives the normal vector at the specified point.

Substituting the given values into the partial derivatives provides the normal vector. Next, we use the point-normal form of the plane equation, where the normal vector is combined with the coordinates of the specified point.

Simplifying this equation yields the final answer \(x - 4y - 3z + 11 = 0\). The resulting equation represents the tangent plane to the parametric surface at the given point.

Answer 2

To find the equation of the tangent plane to the given parametric surface at the specified point, we need to find the partial derivatives, evaluate them at the point, and use them to calculate the normal vector. Finally, we can write the equation of the tangent plane using the formula for a plane and the normal vector.

Step-by-step explanation:

To find the equation of the tangent plane to a parametric surface, we need to compute the partial derivatives and evaluate them at the specified point. Let's start by finding the partial derivatives of the parametric equations:

∂r/∂u = 2ui + 9sinv j + cosv k

∂r/∂v = u² i + 9u cosv j - u sinv k

Next, we'll evaluate these derivatives at u = 2 and v = 0:

∂r/∂u(2,0) = 4i + 9j

∂r/∂v(2,0) = 4i - 2k

Now, we have the normal vector to the tangent plane, which is the cross product of these two vectors:

n = (∂r/∂u) x (∂r/∂v) = (4i + 9j) x (4i - 2k)

Simplifying the cross product gives us n = -18i + 8j + 36k

Finally, we can find the equation of the tangent plane using the formula:

a(x - x0) + b(y - y0) + c(z - z0) = 0,

where (x0, y0, z0) is the given point and (a, b, c) is the normal vector. Plugging in the values, we get:

-18(x - 4) + 8(y - 0) + 36(z - 0) = 0

Simplifying further, the equation of the tangent plane is -18x + 72 + 8y + 36z = 0.