Problem 10 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The equation of the tangent plane is \(z = 2x - y - 1\).

Step 1: Understand the Problem

We need to find the equation of the tangent plane to the surface defined by the function \(f(x, y)=x^{2} e^{-y}\) at the point \((1,0,1)\). A tangent plane to a surface \(z = f(x, y)\) at any point \((x_0, y_0, z_0)\) can be expressed as \(z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\), where \(f_x\) and \(f_y\) are partial derivatives.

Step 2: Compute Partial Derivatives

Calculate the partial derivatives of the function \(f(x, y) = x^2 e^{-y}\). The partial derivative with respect to \(x\) is \(f_x = \frac{\partial}{\partial x}(x^2 e^{-y}) = 2x e^{-y}\), and with respect to \(y\) is \(f_y = \frac{\partial}{\partial y}(x^2 e^{-y}) = -x^2 e^{-y}\).

Step 3: Evaluate Partial Derivatives at Given Point

Evaluate the partial derivatives at the point \((1,0)\). We have \(f_x(1, 0) = 2 \cdot 1 \cdot e^{0} = 2\), and \(f_y(1, 0) = -(1)^2 e^{0} = -1\).

Step 4: Apply the Tangent Plane Formula

Use the tangent plane formula: \(z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\). Substitute \(z_0 = 1\), \(x_0 = 1\), \(y_0 = 0\), \(f_x(1,0) = 2\), and \(f_y(1,0) = -1\). The equation becomes \(z = 1 + 2(x - 1) - 1(y - 0)\).

Step 5: Simplify the Equation

Simplify the equation \(z = 1 + 2(x - 1) - 1y\). This simplifies to \(z = 1 + 2x - 2 - y\), which further simplifies to \(z = 2x - y - 1\).

Key Concepts

Partial Derivatives

When working with functions of two variables, such as the function given in the exercise \(f(x, y) = x^2 e^{-y}\), it's crucial to understand partial derivatives. Partial derivatives measure how the function changes as one of the variables changes while keeping the other constant.

For example, the partial derivative with respect to \(x\), denoted \(f_x\), reflects the rate of change of \(f\) as \(x\) changes but \(y\) remains fixed. Similarly, the partial derivative with respect to \(y\), denoted \(f_y\), measures the change as \(y\) changes while keeping \(x\) constant.

To compute these, perform standard differentiation while treating the other variable as a constant. In this exercise:

• The partial derivative with respect to \(x\) is \(f_x = \frac{\partial}{\partial x}(x^2 e^{-y}) = 2x e^{-y}\).
• The partial derivative with respect to \(y\) is \(f_y = \frac{\partial}{\partial y}(x^2 e^{-y}) = -x^2 e^{-y}\).

Remember that these derivatives are crucial for forming the equation of the tangent plane.

Equation of the Tangent Plane

The equation of a tangent plane to a surface at a given point provides a linear approximation of the surface near that point. For a surface described by \(z = f(x, y)\), the equation of the tangent plane at a point \((x_0, y_0, z_0)\) is:

• \(z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\).

This formula uses the partial derivatives \(f_x\) and \(f_y\), which indicate how the surface slopes along each axis at the point.

In the given exercise, we have:

• The coordinates of the point, \((x_0, y_0, z_0)\) = (1, 0, 1).
• Calculated derivatives: \(f_x(1, 0) = 2\) and \(f_y(1, 0) = -1\).

Plug these values into the tangent plane formula:

\( z = 1 + 2(x - 1) - 1(y - 0) \).

This represents the equation of the tangent plane at the point \((1, 0, 1)\).

Evaluating Derivatives

Evaluating derivatives at a specific point is a key step in finding the equation of the tangent plane. Once you've calculated the partial derivatives generally, substituting the point's coordinates into these derivatives gives the necessary slope values for the tangent plane equation.

In this exercise, after calculating the partial derivatives:

• \(f_x(x, y) = 2x e^{-y}\)
• \(f_y(x, y) = -x^2 e^{-y}\)

we need to evaluate them at the given point \((1,0)\).

By substituting \(x = 1\) and \(y = 0\) into \(f_x\) and \(f_y\):

• \(f_x(1, 0) = 2 \cdot 1 \cdot e^0 = 2\)
• \(f_y(1, 0) = -(1^2) \cdot e^0 = -1\)

This substitution step is essential, as it provides the exact slopes used in the tangent plane equation, giving us the slope and directionality at the specific point on the surface.