Problem 10 - Calculus Early Transcendentals

Question

Use Theorem 7 to find the following derivatives. When feasible, express your answer in terms of the independent variable. $$d z / d t, \text { where } z=x^{2} y-x y^{3}, x=t^{2}, \text { and } y=t^{-2}$$

Step-by-Step Solution

Answer

The derivative of z with respect to t can be found using the chain rule and is given by the expression: \(\frac{dz}{dt} = 2t + 6 - 2t^{-3}\).

Step 1: Compute partial derivatives of z with respect to x and y

First, compute the partial derivatives of z with respect to x and y: $$\frac{\partial z}{\partial x} = 2xy - y^{3}\text{ and } \frac{\partial z}{\partial y} = x^{2} - 3xy^{2}$$

Step 2: Find the derivatives of x and y with respect to t

We also need to find the derivative of x and y with respect to t: $$\frac{dx}{dt} = 2t\text{ and } \frac{dy}{dt} = -2t^{-3}$$

Step 3: Apply The Chain Rule (Theorem 7) to find dz/dt

Now we apply the chain rule to compute \(\frac{dz}{dt}\) using the computed partial derivatives and the derivatives of x and y with respect to t: $$\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$$ Plug in the values from Steps 1 and 2: $$\frac{dz}{dt} = (2xy - y^{3})(2t) + (x^{2} - 3xy^{2})(-2t^{-3})$$

Step 4: Replace x and y with their expressions in terms of t

Replace x and y with their expressions in terms of t, as given in the exercise (\(x=t^2\), \(y=t^{-2}\)): $$\frac{dz}{dt} = (2t^{2}t^{-2} - t^{-6})(2t) + (t^{4} - 3t^{2}t^{-4})(-2t^{-3})$$

Step 5: Simplify the expression for dz/dt

Now, we simplify the expression for \(\frac{dz}{dt}\): $$\frac{dz}{dt} = (2 - t^{-4})(2t) + (t^{4} - 3t^{-2})(-2t^{-3})$$ $$\frac{dz}{dt} = 4t - 2t^{-3} - 2t + 6$$

Step 6: Combine like terms

Combine like terms to get the final expression for \(\frac{dz}{dt}\): $$\frac{dz}{dt} = (4t - 2t) + (6 - 2t^{-3})$$ $$\frac{dz}{dt} = 2t + 6 - 2t^{-3}$$

Key Concepts

Chain Rule

The chain rule is a fundamental principle in calculus that is used to differentiate composite functions. It allows you to break down complex expressions, making it easier to find derivatives. In essence, the chain rule relates the derivative of a composite function to the derivatives of its component functions. When you have a function such as \( z(x, y) \), where both \( x \) and \( y \) are also functions of another variable \( t \), the chain rule helps you find \( \frac{dz}{dt} \).

To apply the chain rule, follow these steps:

• Calculate the partial derivatives of \( z \) with respect to \( x \) and \( y \).
• Determine the derivatives of \( x \) and \( y \) with respect to \( t \).
• Use the chain rule formula: \( \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt} \).
• Substitute the values you've calculated into the formula and simplify.

These steps might seem intricate, but they beautifully decompose the differentiation process, especially when dealing with multiple variables. Remember, practice is key to mastering the chain rule!

Differentiation

Differentiation is one of the core operations in calculus, primarily focusing on finding the rate at which a function is changing at any given point. It deals with the concept of a derivative, which is a measure of how a function's output value changes as its input changes. By differentiating, you can gain insights into the behavior and properties of functions.

The differentiation process involves the following steps:

• Identify the function you need to differentiate.
• Apply rules of differentiation, such as the power rule, product rule, or chain rule where appropriate.
• Simplify the expression obtained after differentiation.

Differentiation can be applied to a wide range of functions including polynomials, exponentials, logarithms, and trigonometric functions. For instance, differentiating \( x = t^2 \) gives \( \frac{dx}{dt} = 2t \), and differentiating \( y = t^{-2} \) results in \( \frac{dy}{dt} = -2t^{-3} \). These derivatives represent the changes in \( x \) and \( y \) concerning \( t \), crucial for applying the chain rule.

Calculus

Calculus is a branch of mathematics that studies how things change. It provides the framework required to analyze change in mathematical models. It is divided into two main parts: differentiation and integration.

Differentiation, as discussed, helps you find the rate of change which is essential in understanding phenomena such as speed, acceleration, and growth rates. Integration, on the other hand, deals with aggregating or totaling pieces to find a whole, such as finding an area under a curve. Together, these concepts allow us to solve many practical problems in science and engineering.

Some key points about calculus include:

• It is used to find maximum and minimum values of functions.
• Helps in understanding the motion and dynamics of systems.
• Allows the computation of areas and volumes of various geometrical shapes.

By mastering calculus, you unlock the ability to tackle complex equations and model real-world systems effectively. The applications are immense, making calculus a valuable tool in both academic studies and various professions.