Question

Tangent Plane In Exercises 31-34, find an equation of the tangent plane to the surface represented by the vector-valued function at the given point. $$ \mathbf{r}(u, v)=(u+v) \mathbf{i}+(u-v) \mathbf{j}+v \mathbf{k}, \quad(1,-1,1) $$

Short answer

The equation of the tangent plane is \(-x + y + z = 1\).

Step-by-step solution

Compute partial derivatives

The first step involves calculating the partial derivatives of the given vector-valued function. The vector-valued function in this case is \[ \mathbf{r}(u, v)=(u+v) \mathbf{i}+(u-v) \mathbf{j}+v \mathbf{k} \]. Let's find the partial derivatives with respect to \(u\) and \(v\). \( \mathbf{r_u} = \frac{\partial \mathbf{r}}{\partial u} = \mathbf{i}+\mathbf{j} \) and \( \mathbf{r_v} = \frac{\partial \mathbf{r}}{\partial v} = \mathbf{i}-\mathbf{j} +\mathbf{k} \).

Compute the normal vector

The normal vector to the surface at the given point can be found by taking the cross product of the partial derivatives, i.e., \( \mathbf{n} = \mathbf{r_u} \times \mathbf{r_v} \). After carrying out the vector product operation, we find that \( \mathbf{n} = -\mathbf{i} -\mathbf{k} \).

Write down the equation of the tangent plane

The equation of a plane given a point \((x_0, y_0, z_0)\) and a normal vector \((a, b, c)\) is given by \(a(x -x_0) + b(y - y_0) + c(z - z_0) = 0 \). Using this form, let's substitute for our point \((1,-1,1)\) and our normal vector \((-1,-1)\). Hence, the equation of the tangent plane is \(-1(x - 1) - 1*(y + 1) - 1*(z - 1) = 0\), which simplifies to \(-x + y + z = 1\).