Question

Tangent Plane, find an equation of the tangent plane to the surface represented by the vector-valued function at the given point. $$ \mathbf{r}(u, v)=u \mathbf{i}+v \mathbf{j}+\sqrt{u v} \mathbf{k}, \quad(1,1,1) $$

Short answer

The equation of the tangent plane to the surface at the point (1,1,1) is \( -x - y + 4z = 2 \) .

Step-by-step solution

Calculate the Partial Derivatives

First, find the partial derivatives of the vector-valued function \(\mathbf{r}(u, v) = u \mathbf{i}+v \mathbf{j}+\sqrt{u v} \mathbf{k}\) with respect to \(u\) and \(v\). For \(u\) it is \( \mathbf{r}_u = \mathbf{i} + 0.5\sqrt{v/u}\mathbf{k}\), and for \(v\) it is \( \mathbf{r}_v = \mathbf{j} + 0.5\sqrt{u/v}\mathbf{k}\).

Evaluate the Derivatives at the Given Point

Plug in the given point \((1,1,1)\) into the derivatives we just found. For \(u\) it becomes \(\mathbf{r}_u(1,1) = \mathbf{i} + 0.5\mathbf{k}\), and for \(v\) it becomes \(\mathbf{r}_v(1,1) = \mathbf{j} + 0.5\mathbf{k}\).

Find the Normal Vector of the Tangent Plane

To find the normal vector to the surface at a given point, we need to take the cross product of these partial derivative vectors. \( \mathbf{N} = \mathbf{r}_u \times \mathbf{r}_v = -0.25\mathbf{i} - 0.25\mathbf{j} + \mathbf{k}\). This vector will be perpendicular to the tangent plane at the point.

Formulate the Equation of the Tangent Plane

Now we are able to write down the equation of the tangent plane. The equation of a plane given a point \( (x_0, y_0, z_0) \) on the plane and a normal vector \( \mathbf{N}=(A,B,C) \) is \( A(x-x_0) + B(y-y_0) + C(z-z_0) = 0 \). Therefore, we get the equation of the tangent plane as \( -0.25(x-1) - 0.25(y-1) + (z-1) = 0 \) or simpler \( -x - y + 4z = 2 \) .