Question

Tangent Plane, find an equation of the tangent plane to the surface represented by the vector-valued function at the given point. $$ \mathbf{r}(u, v)=2 u \cosh v \mathbf{i}+2 u \sinh v \mathbf{j}+\frac{1}{2} u^{2} \mathbf{k}, \quad(-4,0,2) $$

Short answer

The equation of the tangent plane to the surface at the given point (-4,0,2) is \( -4(x+4) + 4y + 16(z-2) = 0 \).

Step-by-step solution

Find the (u,v)

First, find the values of \(u\) and \(v\) corresponding to the point (-4,0,2). From the given position vector \( \mathbf{r}\), we have the k-component as \(1/2 u^{2}\). From this equation and the z-component of the given point, we have \( u = \pm 2 \). Using the i-component of the given point and choosing the negative root (-2) for \(u\) (since the given x-coordinate is negative), we find that \( v = \cosh^{-1}{(-2)} \). But this is not possible as the hyperbolic cosine function always returns a positive value. So we must choose the positive root for \(u\). Therefore, \(u=2\) and \( v = \cosh^{-1}(2) = 0 \)

Calculate Partial Derivatives

Next, calculate the partial derivatives of the given surface vector function \(\mathbf{r}(u, v)\) with respect to both variables \(u\) and \(v\). The resulting vectors are: \( \mathbf{r}_u(u, v) = 2 \cosh{v} \mathbf{i}+ 2 \sinh{v} \mathbf{j}+ u\mathbf{k} \) and \(\mathbf{r}_v(u, v) = 2u \sinh{v} \mathbf{i}+ 2u \cosh{v} \mathbf{j} \)

Find the Cross Product

Calculate the cross product of the two partial derivatives \(\mathbf{r}_u(u, v)\) and \(\mathbf{r}_v(u, v)\). This will give us a normal vector to the tangent plane. \( \mathbf{N} = \mathbf{r}_u \times \mathbf{r}_v = -u^{2}\mathbf{i} + u^{2}\mathbf{j} + 4u^{2}\mathbf{k} \)

Find the Plane Equation

Using the normal vector and the given point (-4,0,2), find the equation of the tangent plane in the standard form \((a(x-x_0) + b(y-y_0) + c(z-z_0) = 0)\), where (a,b,c) are the components of the normal vector and (x_0,y_0,z_0) is the point on the plane. Substituting the values we get the equation to be \( -4(x+4) + 4y + 16(z-2) = 0 \)

Key concepts

Vector-Valued Functions

A vector-valued function is a function that takes one or more variables and returns a vector. In the context of the given exercise, the vector-valued function is represented by \( \mathbf{r}(u, v) \) and maps two parameters, \( u \) and \( v \) into a three-dimensional space. The components of the resulting vector are functions of the parameters, often involving operations like trigonometric functions or polynomials.

Understanding vector-valued functions is crucial because these functions can represent curves or surfaces in space. For example, in the given problem, the function \( \mathbf{r}(u, v) \) describes a surface. The ability to identify and interpret the specific surface is important in various fields of science and engineering such as physics, graphics design, and calculus.

Partial Derivatives

Partial derivatives are foundational in multivariable calculus. When working with functions of more than one variable, like vector-valued functions, we often want to understand how the function changes with respect to one variable while holding the others constant. These rates of change are captured by partial derivatives.

Calculating partial derivatives is a step-by-step process that requires treating all other variables as constants while differentiating with respect to the variable of interest. In the context of our tangent plane problem, calculating partial derivatives of the vector-valued function \( \mathbf{r}(u, v) \) with respect to \( u \) and \( v \) separately provides us with two new vectors that are tangential to the surface at a given point. These vectors are instrumental in determining the plane tangent to the surface at that point.

Cross Product

The cross product, also known as the vector product, is a binary operation on two vectors in three-dimensional space. It results in a third vector that is perpendicular to the plane containing the original vectors. The magnitude of the cross product vector is equal to the area of the parallelogram that the two vectors span. One common use for the cross product is to find a normal vector to a plane.

In the problem at hand, the cross product of the partial derivative vectors \( \mathbf{r}_u \) and \( \mathbf{r}_v \) provides us with a normal vector to the surface at the point of interest. This normal vector is essential for writing the equation of the tangent plane because it defines the orientation of the plane in space. The components of this normal vector become the coefficients in the standard form equation of the plane, ensuring that the plane is indeed tangent to the given surface.