Question

Find the equation of the tangent plane to the surface $$z=f(x, y) \quad \text { at } \quad\left(x_{0}, y_{0}, z_{0}\right)=\left(x_{0}, y_{0}, f\left(x_{0}, y_{0}\right)\right)$$ (a) \(f(x, y)=x^{2}+y^{2}-1, \quad\left(x_{0}, y_{0}\right)=(1,2)\) (b) \(f(x, y)=2 x+3 y+1, \quad\left(x_{0}, y_{0}\right)=(1,-1)\) (c) \(f(x, y)=x y \sin x y, \quad\left(x_{0}, y_{0}\right)=(1, \pi / 2)\) (d) \(f(x, y)=x^{2}-2 y^{2}+3 x y, \quad\left(x_{0}, y_{0}\right)=(2,-1)\)

Short answer

a) Given function: \(f(x, y) = x^{2} + y^{2} - 1 \), and point (1, 2) Answer: \(2x(x-1) + 2y(y-2) - z + 3 = 0\) b) Given function: \( f(x, y) = x - 2y\), and point (3, 1) Answer: \(x - 2y - z + 6 = 0\) c) Given function: \(f(x, y) = \pi x \cos(\frac{\pi}{2}) + \pi y\), and point (\(\frac{\pi}{2}\), \(\pi\)) Answer: \(\pi(y - \pi\cos(\pi/2) - z) = 0\) d) Given function: \(f(x, y) = x^3 + 2x^{2}y - 2y^{3} + y^2\), and point (2, 1) Answer: \(3x^{2} + 4xy - 4y^{2} + y - z - 6 = 0\)

Step-by-step solution

Find the partial derivatives of the function

We have the function: $$ f(x, y) = x^{2}+y^{2}-1 $$ Compute the partial derivatives with respect to x and y: $$ \frac{\partial f}{\partial x} = 2x, \quad \frac{\partial f}{\partial y} = 2y $$

Find the point on the surface

For (x₀, y₀) = (1, 2), find z₀ = f(x₀, y₀) as follows: $$ z_{0} = f(1, 2) = 1^{2} + 2^{2} - 1 = 0 + 4 - 1 = 3 $$ Hence, the point on the surface is (1, 2, 3).

Calculate the tangent plane equation

For the tangent plane equation, substitute x₀, y₀, z₀, ∂f/∂x, and ∂f/∂y: $$ (2x) * (x - 1) + (2y) * (y - 2) - (z - 3) = 0 \\ 2x(x-1) + 2y(y-2) - z + 3 = 0 $$ The tangent plane equation for case (a) is: $$ 2x(x-1) + 2y(y-2) - z + 3 = 0 $$ For (b), (c), and (d) the process will be the same: first calculate the partial derivatives of the given functions, then determine the point on the surface, and lastly derive the tangent plane equations by substituting in the given parameters. The equations will be obtained as: (b) $$x - 2y - z + 6 = 0 $$ (c) $$\pi(y - \pi\cos(\pi/2) - z) = 0$$ (d) $$3x^{2} + 4xy - 4y^{2} + y - z - 6 = 0$$

Key concepts

Partial Derivatives

Understanding partial derivatives is essential in multivariable calculus, where functions have more than one variable. In such functions, a partial derivative represents how the function changes as one variable changes, while others are held constant. For example, if you have a function,
\( f(x, y) = x^2 + y^2 \),
to find how the function changes in the x-direction at any point, we take the partial derivative with respect to x:
\( \frac{\partial f}{\partial x} = 2x \).
Similarly, the rate of change in the y-direction is:
\( \frac{\partial f}{\partial y} = 2y \).
These partial derivatives are the foundational building blocks for finding the slope of the tangent plane to a surface at a given point. The process involves evaluating these derivatives at specific points to define the plane uniquely.

Multivariable Calculus

In multivariable calculus, we deal with functions of several variables and examine changes in different directions. This field of mathematics extends the principles of calculus to multiple dimensions. An integral concept within this domain is the 'tangent plane to a surface.' It represents a plane that just touches a surface at a given point and has the same slope as the surface at that point. To find this plane, you first need the gradient at the point of tangency, which is obtained from partial derivatives, as was seen in the first example.
By using gradients and directional derivatives, multivariable calculus allows us to analyze and solve problems related to 3D shapes and how they change.

Tangent Plane to Surface

The equation of a tangent plane to a given surface at a specific point is crucial in understanding the immediate behavior of that surface. To find this plane, we determine the point of contact and the surface's slope at that point using partial derivatives. If we have a function
\( z = f(x, y) \),
and a point on the surface
\( (x_0, y_0, z_0) \),
the equation of the tangent plane at that point can be derived as
\((\frac{\partial f}{\partial x} \bigg\vert _{x_0,y_0})(x - x_0) + (\frac{\partial f}{\partial y} \bigg\vert _{x_0,y_0})(y - y_0) - (z - z_0) = 0 \).
This equation relates all three coordinates, establishing a plane that locally approximates the surface.

Gradient Vector

The gradient vector plays a significant role in finding the tangent plane to a surface. It combines all partial derivatives of a function into a vector that points in the direction of the greatest rate of increase of the function. The gradient is denoted as grad(f) or
\( abla f \),
and in two variables, it's formed as
\( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).
At a point
\( (x_0, y_0) \),
the gradient is evaluated as
\( abla f(x_0, y_0) \).
This vector is essential for the tangent plane equation because its components give the coefficients of x and y in the equation.
The gradient can also indicate the 'steepness' and direction in which the surface rises most rapidly from the point of tangency. By including the gradient in our understanding of tangent planes, we gain more insights into the behavior of multi-dimensional functions.