Problem 20 - Calculus with Concepts in Calculus

Question

Compute \(d w / d t\) $$ w=\sqrt{x^{2}+y^{2}}-\sqrt{y^{3}-z^{3}} ; x=t^{2}, y=t^{3}, z=-t^{3} $$

Step-by-Step Solution

Answer

The derivative \( \frac{dw}{dt} \) is 0.

Step 1: Understand the Function

The function given is \( w = \sqrt{x^2 + y^2} - \sqrt{y^3 - z^3}\). Here, \( x = t^2 \), \( y = t^3 \), and \( z = -t^3 \). We need to find \( \frac{dw}{dt} \) using these expressions for \(x\), \(y\), and \(z\).

Step 2: Find Partial Derivatives with Respect to x, y, and z

To find \( \frac{dw}{dt} \), it is helpful first to compute partial derivatives. Find: \( \frac{\partial w}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} \), \( \frac{\partial w}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} - \frac{3y^2}{2\sqrt{y^3 - z^3}} \), \( \frac{\partial w}{\partial z} = \frac{3z^2}{2\sqrt{y^3 - z^3}} \).

Step 3: Apply Chain Rule

Using the chain rule, we have:\[ \frac{dw}{dt} = \frac{\partial w}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial w}{\partial y} \cdot \frac{dy}{dt} + \frac{\partial w}{\partial z} \cdot \frac{dz}{dt} \]Substitute \( \frac{dx}{dt} = 2t \), \( \frac{dy}{dt} = 3t^2 \), and \( \frac{dz}{dt} = -3t^2 \).

Step 4: Substitute and Simplify

Substitute \( x = t^2 \), \( y = t^3 \), \( z = -t^3 \) and their derivatives into the expression:\[ \frac{dw}{dt} = \frac{t^2}{\sqrt{t^4 + t^6}} \cdot 2t + \left(\frac{t^3}{\sqrt{t^4 + t^6}} - \frac{3(t^3)^2}{2\sqrt{t^9 - (-t^3)^3}}\right) \cdot 3t^2 + \frac{3(-t^3)^2}{2\sqrt{t^9 - (-t^3)^3}} \cdot (-3t^2) \].

Step 5: Simplify Further

Further simplify by evaluating each component:\( \frac{t^2}{\sqrt{t^4 + t^6}} = \frac{1}{t} \),The terms within the parentheses become zero because the terms cancel each other, simplifying to zero since \( y^3 = z^3 \).Thus, \( \frac{dw}{dt} \) simplifies to zero.

Step 6: Conclusion

The derivative \( \frac{dw}{dt} \) simplifies to 0 because the terms cancel each other out.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes. It is used to calculate the derivative of a function, which represents the slope of the function at any given point. Knowing this can help determine how the function behaves. For example, in physics, it can tell us how the velocity of an object changes over time. Differentiation is key when examining how various quantities relate to one another and change. It provides insights into both linear and non-linear relationships.

• The general formula to find the derivative of a function, say \( f(x) \), is \( f'(x) \) or \( \frac{df}{dx} \).
• It finds applications in various fields like physics, engineering, biology, and economics.

Understanding differentiation allows us to tackle more complex problems, like those involving functions of multiple variables.

Chain Rule

The chain rule is a powerful method in calculus used for finding the derivative of composite functions. It provides a systematic way to differentiate when functions are nested within each other. When you have a function within another function, the chain rule helps you find the overall derivative by focusing on the inner and outer functions separately.

Here's the basic idea: if you have a composite function \( f(g(x)) \), the chain rule states that its derivative is \( f'(g(x)) \cdot g'(x) \). It breaks down the differentiation into manageable steps for complex expressions.

• The outer function is differentiated first, treating the inner function as a simple variable.
• Then multiply by the derivative of the inner function.

In many problems, like the one given, utilizing the chain rule is crucial for finding the rate of change with respect to time or any other variables.

Partial Derivatives

Partial derivatives involve differentiation of functions with multiple variables, focusing on one variable at a time while keeping others constant. This concept is essential in multivariable calculus, where functions depend on two or more variables. It helps us understand how a change in one variable affects the overall function.

For a function \( f(x, y, z) \), the partial derivative with respect to \( x \) is denoted by \( \frac{\partial f}{\partial x} \). Similarly, partial derivatives can be taken with respect to \( y \) and \( z \). By calculating these, we can explore how each variable individually influences the function.

• They are key in tasks such as optimizing functions or solving equations in physics and engineering.
• Partial derivatives are also the building blocks for understanding gradient vectors and surface geometry.

In problems involving differentiation, partial derivatives help break down complex relationships to find rates of change for each independent variable, setting up the use of the chain rule effectively.