Problem 9 - APEX Calculus

Question

In Exercises \(7-12,\) functions \(z=f(x, y), x=g(t)\) and \(y=h(t)\) are given. (a) Use the Multivariable Chain Rule to compute \(\frac{d z}{d t}\). (b) Evaluate \(\frac{d z}{d t}\) at the indicated \(t\) -value. $$ z=5 x+2 y, \quad x=2 \cos t+1, \quad y=\sin t-3 ; \quad t=\pi / 4 $$

Step-by-Step Solution

Answer

\( \frac{dz}{dt} \) at \( t=\frac{\pi}{4} \) is \(-4\sqrt{2}\).

Step 1: Identify Given Functions

In the problem, we have three functions: \( z = f(x, y) = 5x + 2y \), \( x = g(t) = 2\cos t + 1 \), and \( y = h(t) = \sin t - 3 \). These functions will allow us to find \( \frac{dz}{dt} \) using the chain rule.

Step 2: Apply the Multivariable Chain Rule

The Multivariable Chain Rule for \( z = f(x, y) \) is given by \( \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt} \). We need to find \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \), and compute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

Step 3: Find Partial Derivatives

Calculate \( \frac{\partial z}{\partial x} = 5 \) and \( \frac{\partial z}{\partial y} = 2 \), as these are the coefficients from the function \( z = 5x + 2y \).

Step 4: Compute Derivatives of x and y with respect to t

Differentiate the functions: \( \frac{dx}{dt} = \frac{d}{dt}(2\cos t + 1) = -2\sin t \) and \( \frac{dy}{dt} = \frac{d}{dt}(\sin t - 3) = \cos t \).

Step 5: Substitute and Compute \( \frac{dz}{dt} \)

Substitute the derivatives back into the chain rule formula: \( \frac{dz}{dt} = 5(-2\sin t) + 2(\cos t) \). Simplify this to get \( \frac{dz}{dt} = -10\sin t + 2\cos t \).

Step 6: Evaluate at \( t = \frac{\pi}{4} \)

Substitute \( t = \frac{\pi}{4} \) into the expression: Since \( \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \) and \( \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \), \( \frac{dz}{dt} = -10\left(\frac{\sqrt{2}}{2}\right) + 2\left(\frac{\sqrt{2}}{2}\right) \). This simplifies to \( \frac{dz}{dt} = -4\sqrt{2} \).

Key Concepts

Multivariable Chain Rule

In multivariable calculus, the Chain Rule is a fundamental tool that helps differentiate compositions of functions. When dealing with functions of several variables, like in our exercise, the Multivariable Chain Rule becomes especially powerful. It relates the rate of change of a function, depending on multiple variables, to the rate of change of those variables themselves.
For a function like \( z = f(x, y) \) where \( x \) and \( y \) are both functions of \( t \), the Chain Rule allows us to find \( \frac{dz}{dt} \). This is the total derivative of \( z \) with respect to \( t \).
The formula is:

• \( \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt} \)

This formula tells us to:

• First, find the partial derivative of \( z \) with respect to each independent variable \( x \) and \( y \).
• Then, multiply these partials by the derivatives of those variables with respect to \( t \).

Finally, sum these products to get \( \frac{dz}{dt} \), showing how \( z \) changes over time.

Partial Derivatives

Partial derivatives are central to understanding functions of multiple variables. When you have a function like \( z = f(x, y) = 5x + 2y \), and you want to examine how \( z \) changes when one variable changes, you use partial derivatives.

• The partial derivative with respect to \( x \), denoted as \( \frac{\partial z}{\partial x} \), shows how \( z \) changes with \( x \) while holding \( y \) constant.
• The partial derivative with respect to \( y \), \( \frac{\partial z}{\partial y} \), shows how \( z \) changes with \( y \) while holding \( x \) constant.

In our solution, we found these partial derivatives for \( z = 5x + 2y \) as follows:

• \( \frac{\partial z}{\partial x} = 5 \)
• \( \frac{\partial z}{\partial y} = 2 \)

These values indicate the sensitivity of \( z \) to changes in \( x \) and \( y \). Multiplying these partials with derivatives of \( x \) and \( y \) with respect to \( t \) gives insights into the overall behavior of \( z \) in a dynamic system.

Parametric Equations

Parametric equations often represent curves by defining coordinates as functions of a parameter. In this exercise, \( x = 2\cos t + 1 \) and \( y = \sin t - 3 \) are parametric equations.
These describe a path in the \( xy \)-plane as \( t \) varies, typically representing time. \( x \) and \( y \) change based on \( t \), which is our parameter.
Understanding parametric equations is essential because:

• They allow you to describe motion and paths that are not easily expressed as functions \( y = f(x) \).
• They enable calculation of rates of change along paths, using derivatives like \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

In our problem, derivatives of these parametric functions represent the change in \( x \) and \( y \) as \( t \) changes, which fuels the application of the Multivariable Chain Rule efficiently.

Derivative Evaluation

After finding expressions for derivatives using the Chain Rule, the final step is to evaluate these at a specific point. In our exercise, we needed to evaluate \( \frac{dz}{dt} \) at \( t = \frac{\pi}{4} \).
This involves replacing \( t \) with the given value in the derivative expression:

• First, substitute \( t = \frac{\pi}{4} \) into the functions \( \sin t = \frac{\sqrt{2}}{2} \) and \( \cos t = \frac{\sqrt{2}}{2} \).
• Replace these in the solved derivative \( \frac{dz}{dt} = -10\sin t + 2\cos t \).

Calculations yield:

• \( \frac{dz}{dt} = -10\left(\frac{\sqrt{2}}{2}\right) + 2\left(\frac{\sqrt{2}}{2}\right) \)
• Simplifying gives \( -4\sqrt{2} \)

Evaluating derivatives helps us understand how a function behaves at specific instances, turning mathematical maps into tangible insights about the system's evolution.