Problem 2 - 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. \(f(x, y)=x^{2}-3 y^{2} ;(1,2,-11)\)

Step-by-Step Solution

Answer

The equation of the tangent plane is \( z = 2x - 12y + 11 \).

Step 1: Find Partial Derivatives

To find the equation of the tangent plane, we need the partial derivatives of the function with respect to both \(x\) and \(y\). The function is given by \(f(x, y) = x^2 - 3y^2\). First, calculate the partial derivative with respect to \(x\):\[ f_x(x, y) = \frac{\partial}{\partial x}(x^2 - 3y^2) = 2x. \] Next, calculate the partial derivative with respect to \(y\):\[ f_y(x, y) = \frac{\partial}{\partial y}(x^2 - 3y^2) = -6y. \]

Step 2: Evaluate Partial Derivatives at (1, 2)

Substitute \((x_0, y_0) = (1, 2)\) into the partial derivatives obtained from Step 1. Evaluate the partial derivative with respect to \(x\) at the point (1, 2):\[ f_x(1, 2) = 2(1) = 2. \] Evaluate the partial derivative with respect to \(y\) at the point (1, 2):\[ f_y(1, 2) = -6(2) = -12. \]

Step 3: Substitute Values Into the Tangent Plane Formula

The formula for the tangent plane is given by: \[ z = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0). \] We know \(f(x_0, y_0) = f(1, 2) = (1)^2 - 3(2)^2 = 1 - 12 = -11\). Substitute all known values: \[ z = -11 + 2(x - 1) - 12(y - 2). \] Simplify: \[ z = -11 + 2x - 2 - 12y + 24. \] \[ z = 2x - 12y + 11. \]

Step 4: Write Final Equation of the Tangent Plane

The simplified equation from Step 3 is the equation of the tangent plane at the point \((1, 2, -11)\):\[ z = 2x - 12y + 11. \]

Key Concepts

Partial Derivatives

When we deal with functions of multiple variables, like in the example function \( f(x, y) = x^2 - 3y^2 \), we use partial derivatives to measure how the function changes as one of the variables changes, keeping the others constant. In simpler terms, it's a way to see how the function behaves if we tweak one variable while ignoring the others for the moment.

• Partial derivative with respect to \( x \): Here, we treat \( y \) as a constant and differentiate \( f(x, y) \) with respect to \( x \). For our function, that means taking the derivative of \( x^2 \), which gives us \( 2x \), while the \( -3y^2 \) part disappears because it's independent of \( x \).


• Partial derivative with respect to \( y \): Similarly, when we take the partial derivative with respect to \( y \), we treat \( x \) as a constant. So, the derivative of \( -3y^2 \) results in \( -6y \), and the \( x^2 \) part vanishes here.

Understanding partial derivatives is crucial because they form the building blocks for finding the tangent plane and further exploring functions of several variables.

Function Evaluation

Function evaluation is the process of finding the output of a function for specific inputs. It helps us understand specific characteristics of the function at certain points. In our exercise, we evaluated the function \( f(x, y) = x^2 - 3y^2 \) at the point \((1, 2)\).

• Substitute \((x, y) = (1, 2)\) into the function:

This means replacing \( x \) with 1 and \( y \) with 2.

• This gives us: \( f(1, 2) = 1^2 - 3(2)^2 = 1 - 12 = -11 \).

Evaluating a function at a certain point gives us the exact "height" or value of the function at that particular point in its domain. Here, \(-11\) is the height of the graph at the point \((1,2)\), and it's crucial for finding the equation of the tangent plane.

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions with more than one variable. It's a powerful tool used in various fields such as physics, engineering, and economics.

• Functions of multiple variables: These functions help model phenomena that depend on several factors. For instance, the shape of a landscape might depend on various coordinates \((x, y)\).
• Tangent planes: While in single-variable calculus we have tangent lines, in multivariable calculus, we have tangent planes that can touch a surface at only one point, providing the best linear approximation there.

In our case, the tangent plane equation \( z = 2x - 12y + 11 \) was derived using the concepts from multivariable calculus, helping us approximate the surface near the point \((1, 2, -11)\). Understanding these principles is essential for tackling more complex scenarios in higher dimensions.

Gradient

The gradient is a vector that represents the direction and rate of the fastest increase of a function. In simpler terms, it tells us which way to move to make the function grow the quickest.

• Components of the gradient: The gradient vector for a function \( f(x, y) \) combines all its partial derivatives. It's represented as \[ abla f = \left( f_x, f_y \right) \]. For our function, this becomes \[ abla f = (2x, -6y) \].
• Evaluated gradient: At a particular point like \((1, 2)\), it gets specific values: \[ (2\times1, -6\times2) = (2, -12) \]. This tells us that at this point, the function increases fastest in the direction of this vector.

The gradient is fundamental in multivariable calculus because it helps us understand not just how a function behaves, but also where it's heading within its domain. It played a key role in determining the tangent plane by providing the necessary slopes.