Problem 6 - Calculus for Biology and Medicine

Question

The tangent plane at the indicated point \(\left(x_{0}, y_{0}, z_{0}\right)\) exists. Find its equation. $$ f(x, y)=e^{x-y} ;\left(1,-1, e^{2}\right) $$

Step-by-Step Solution

Answer

The tangent plane equation is \(z = e^2x - e^2y\).

Step 1: Determine Partial Derivatives

To find the equation of the tangent plane, we first need the partial derivatives \(f_x(x, y)\) and \(f_y(x, y)\) of the function \(f(x, y) = e^{x-y}\). These derivatives are found using the basic rules of differentiation.- The partial derivative with respect to \(x\) is: \(f_x(x, y) = \frac{d}{dx} e^{x-y} = e^{x-y}\).- The partial derivative with respect to \(y\) is: \(f_y(x, y) = \frac{d}{dy} e^{x-y} = -e^{x-y}\).

Step 2: Evaluate Partial Derivatives at Given Point

Next, we evaluate the partial derivatives at the point \((x_0, y_0) = (1, -1)\).- \(f_x(1, -1) = e^{1 - (-1)} = e^{2}\)- \(f_y(1, -1) = -e^{1 - (-1)} = -e^{2}\)

Step 3: Write the Tangent Plane Equation

The equation of the tangent plane at the point \((x_0, y_0, z_0) = (1, -1, e^2)\) is given by:\[ z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) \]Substituting the evaluated partial derivatives and given point:\[ z - e^2 = e^2(x - 1) - e^2(y + 1) \]

Step 4: Simplify the Tangent Plane Equation

Now simplify the equation from Step 3:\[ z - e^2 = e^2(x - 1) - e^2(y + 1) \]This expands to:\[ z - e^2 = e^2x - e^2 - e^2y - e^2 \]Simplify further:\[ z = e^2x - e^2y \]

Key Concepts

Partial Derivatives

In calculus, partial derivatives are a fundamental tool used when dealing with multivariable functions, like in our task of finding a tangent plane to a surface. A partial derivative represents how a function changes as one of its variables changes, while the others are held constant. For example, with the function given in the exercise, \( f(x,y) = e^{x-y} \), the partial derivative with respect to \( x \), denoted as \( f_x(x, y) \), measures how \( f \) changes as \( x \) changes, keeping \( y \) fixed. Similarly, \( f_y(x, y) \) looks at the changes with respect to \( y \).

• Partial derivatives are crucial for understanding the slope and behavior of multivariable functions.
• They are typically calculated using differentiation rules, treating all other variables as constants.

By determining these derivatives first, we can gain insight into how to construct equations like the tangent plane equation seen in the problem.

Differentiation Rules

To find the partial derivatives necessary for the tangent plane equation, one needs to apply specific differentiation rules. The fundamental rules used in this context are similar to those applied in single-variable calculus but adapted for partial application.
**Power Rule and Exponential Rule:**
For a function involving an exponential like \( e^{x-y} \), the exponential rule is a key differentiation tool. When differentiating \( e^{x} \), its derivative remains \( e^{x} \), but we must consider the chain rule when dealing with a composite function like \( e^{x-y} \).

• The derivative with respect to \( x \) treats \( y \) as a constant, resulting in \( e^{x-y} \).
• The derivative with respect to \( y \), recognizing the negative sign, results in \( -e^{x-y} \).

Understanding these rules helps when working through partial derivatives, allowing us to compute slopes in different directions on a surface.

Evaluate Derivatives

After computing partial derivatives, evaluating them at a certain point is a pivotal step in deriving the tangent plane equation. This step involves plugging specific values into the partial derivatives to get the slope at that exact point on the surface.
In our example, we evaluated \( f_x(x, y) = e^{x-y} \) and \( f_y(x, y) = -e^{x-y} \) at the point \((1, -1)\):

• For \( f_x(1, -1) \), substitute \( x = 1 \) and \( y = -1 \) into \( e^{x-y} \) to get \( e^{2} \).
• For \( f_y(1, -1) \), substitute the same values to get \( -e^{2} \).

These evaluations give us the coefficients for the linear terms in the tangent plane equation. With these values, the tangent plane can be expressed, providing a linear approximation to the surface at the point of interest.