Question

Looking ahead- tangent planes Consider the following surfaces \(f(x, y, z)=0,\) which may be regarded as a level surface of the function \(w=f(x, y, z) .\) A point \(P(a, b, c)\) on the surface is also given. a. Find the (three-dimensional) gradient of \(f\) and evaluate it at \(P\). b. The set of all vectors orthogonal to the gradient with their tails at \(P\) form a plane. Find an equation of that plane (soon to be called the tangent plane). $$f(x, y, z)=e^{x+y-z}-1=0 ; P(1,1,2)$$

Short answer

Question: Find the gradient of the surface defined by the implicit function \(f(x, y, z)= e^{x+y-z} - 1\) at the point \(P(1,1,2)\) and the equation of the tangent plane at that point. Answer: The gradient vector at the point \(P(1,1,2)\) is given by \(\nabla f(P) = (1, 1, -1)\). The equation of the tangent plane at that point is \(x + y - z = 0\).

Step-by-step solution

Calculate the gradient vector

Calculate the partial derivatives of the function \(f(x, y, z)\): $$\frac{\partial f}{\partial x} = \frac{d}{dx}(e^{x+y-z}-1) = e^{x+y-z}$$ $$\frac{\partial f}{\partial y} = \frac{d}{dy}(e^{x+y-z}-1) = e^{x+y-z}$$ $$\frac{\partial f}{\partial z} = \frac{d}{dz}(e^{x+y-z}-1) = -e^{x+y-z}$$ Thus, the gradient vector of \(f\) is given by: $$\nabla f = \left( e^{x+y-z}, e^{x+y-z}, -e^{x+y-z} \right)$$

Evaluate the gradient at point P

Now we need to find the gradient vector at the point \(P(1,1,2)\). Plug the coordinates of P into the gradient vector: $$\nabla f(P) = \left( e^{1+1-2}, e^{1+1-2}, -e^{1+1-2} \right) = \left( e^0, e^0, -e^0 \right) = \left( 1, 1, -1 \right)$$

Find the equation of tangent plane

To find the equation of the tangent plane orthogonal to the gradient vector at point \(P\), we need to use the dot product of the gradient vector and the difference between a general point in the plane \((x,y,z)\) and \(P(1,1,2)\): $$(x - 1, y - 1, z - 2) \cdot \nabla f(P) = 0$$ This dot product will give us: $$(x - 1) \times 1 + (y - 1) \times 1 + (z - 2) \times (-1) = 0$$ Simplifying the equation, we get the equation for the tangent plane at \(P(1,1,2)\): $$x + y - z = 0$$

Key concepts

Gradient Vector

The gradient vector is a fundamental concept in multivariable calculus. It's essentially a vector field composed of the partial derivatives of a function. In this context, it provides critical information about the direction and rate at which a function changes the fastest.
For a function of three variables, such as \(f(x, y, z)\), the gradient vector \(abla f\) is given as:

• \(abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)\)

In the exercise at hand, this means taking the derivatives of \(f\) with respect to each variable individually while holding the others constant. This results in a vector comprising these partial derivatives.
The gradient is significant because it points perpendicular to the level surfaces at a given point. At the point \(P(a, b, c)\), evaluating the gradient vector, one obtains the specific gradient value at that point. This information is then used to determine the orientation of the tangent plane.

Partial Derivatives

Partial derivatives are key building blocks in multivariable calculus. They represent the rate of change of a function with respect to one of its variables, keeping the other variables constant. In surface analysis, partial derivatives help in understanding how a surface changes.
For a function \(f(x, y, z)\), the partial derivatives are calculated as follows:

• \(\frac{\partial f}{\partial x}\): derivative of \(f\) with respect to \(x\).
• \(\frac{\partial f}{\partial y}\): derivative of \(f\) with respect to \(y\).
• \(\frac{\partial f}{\partial z}\): derivative of \(f\) with respect to \(z\).

Each of these derivatives gives insight into how the function changes in different directions from a point on its surface. Calculating these derivatives allows us to form the gradient vector, which is pivotal in finding the tangent plane.
The exercise provided uses these partial derivatives to form that gradient vector. At point \(P(1, 1, 2)\), the specific values of these derivatives indicate the rate and direction of change, and thus help frame the equation of the plane that just touches the surface at the point.

Equation of Plane

An equation of a plane in space gives a flat surface extending infinitely in two dimensions. In this exercise, we find the equation of a tangent plane to a surface at a given point. This plane is perpendicular (orthogonal) to the gradient vector at that point.
To write this equation:

• Use the formula: \((x - a)\frac{\partial f}{\partial x} + (y - b)\frac{\partial f}{\partial y} + (z - c)\frac{\partial f}{\partial z} = 0\)
• In the exercise, the gradient is \((1, 1, -1)\) at point \(P(1, 1, 2)\).
• This simplifies to \((x - 1) + (y - 1) - (z - 2) = 0\).

Solving this gives the equation of the tangent plane as \(x + y - z = 0\). This plane just "touches" the surface at \(P\), acting as the flat approximation to the surface near the point.
It's important because it provides an approach to local linearizations of functions, where complex surfaces can be approximated by simpler, planar regions.