Problem 10 - 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. $$ z=\frac{x}{y^{2}+1}, \quad x=\cos t, \quad y=\sin t ; \quad t=\pi / 2 $$

Step-by-Step Solution

Answer

\(\frac{dz}{dt}\) is \(-\frac{1}{2}\) at \(t = \pi/2\).

Step 1: Understand the Functions

We are given the function \(z = \frac{x}{y^2 + 1}\), where \(x = \cos t\) and \(y = \sin t\). Our task is to compute \(\frac{dz}{dt}\) using the Chain Rule of multivariable calculus.

Step 2: Identify the Partial Derivatives

Identify the partial derivatives of \(z\) with respect to \(x\) and \(y\). These are: \(\frac{\partial z}{\partial x} = \frac{1}{y^2 + 1}\) and \(\frac{\partial z}{\partial y} = -\frac{2xy}{(y^2 + 1)^2}\).

Step 3: Apply the Multivariable Chain Rule

Use the chain rule formula \(\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}\). Compute \(\frac{dx}{dt} = -\sin t\) and \(\frac{dy}{dt} = \cos t\).

Step 4: Substitute and Simplify

Substitute \( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, \frac{dx}{dt}, \) and \( \frac{dy}{dt} \) into the chain rule equation:\[ \frac{dz}{dt} = \left(\frac{1}{y^2 + 1}\right)(-\sin t) + \left(-\frac{2xy}{(y^2 + 1)^2}\right)(\cos t). \] Insert the expressions for \(x = \cos t\) and \(y = \sin t\).

Step 5: Evaluate at \(t = \frac{\pi}{2}\)

First, compute \(x = \cos\left(\frac{\pi}{2}\right) = 0\) and \(y = \sin\left(\frac{\pi}{2}\right) = 1\). Plug these values into the expression for \(\frac{dz}{dt}\):\[ \frac{dz}{dt} = \left(\frac{1}{1^2 + 1}\right)(-1) + \left(-\frac{2 \cdot 0 \cdot 1}{(1^2 + 1)^2}\right)(0), \] resulting in \(\frac{dz}{dt} = -\frac{1}{2} + 0 = -\frac{1}{2}.\)

Key Concepts

Partial Derivatives

Understanding partial derivatives is crucial when dealing with functions of several variables. In this context, a partial derivative measures how a function changes as one of its variables changes, while the others remain constant.
For a function like \( z = \frac{x}{y^2 + 1} \), we find partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \).

Here's a quick breakdown:

• \( \frac{\partial z}{\partial x} = \frac{1}{y^2 + 1} \): This shows how \( z \) changes as \( x \) changes, keeping \( y \) constant.
• \( \frac{\partial z}{\partial y} = -\frac{2xy}{(y^2 + 1)^2} \): This reflects the change in \( z \) with variations in \( y \), while \( x \) remains fixed.

By recognizing these derivatives, we can understand how changes in each variable will impact the function's outcome.

Chain Rule in Calculus

The chain rule is a fundamental concept in calculus, especially when dealing with composite functions. When used in multivariable calculus, it allows us to find the derivative of a function with several variables in relation to another variable.

To apply the multivariable chain rule for a function \( z = f(x, y) \) where \( x = g(t) \) and \( y = h(t) \):

• We use the formula: \( \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt} \).
• This requires calculating \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), i.e., the derivatives of \( x \) and \( y \) with respect to \( t \).

Understanding this rule is pivotal, as it ties together multiple functions into one unified derivative. This capability is essential for examining how changes in independent variables impact the function being analyzed.

Trigonometric Identities

Trigonometric identities are useful tools that simplify the process of finding derivatives in trigonometric calculations. In many cases, when functions involve \( \sin \) and \( \cos \), knowing these identities can help streamline the process.
For example, consider:

• \( x = \cos(t) \) implies \( \frac{dx}{dt} = -\sin(t) \)
• \( y = \sin(t) \) implies \( \frac{dy}{dt} = \cos(t) \)

Using these identities simplifies the derivative finding process by linking the function's components together.

Additionally, applying these identities at specific points, such as \( t = \frac{\pi}{2} \), helps evaluate the function's behavior at that exact moment. In this case, using \( \cos\left(\frac{\pi}{2}\right) = 0 \) and \( \sin\left(\frac{\pi}{2}\right) = 1 \), we can determine the values needed for accurate differentiation.