Problem 2 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(d w / d t\) by using the Chain Rule. Express your final answer in terms of \(t\). $$ w=x^{2} y-y^{2} x ; x=\cos t, y=\sin t $$

Step-by-Step Solution

Answer

\( \frac{dw}{dt} = \cos t (\cos^2 t - 2\sin^2 t) + \sin t (\sin^2 t - 2\cos^2 t) \)

Step 1: Identify the Variables

The function given is \(w = x^2y - y^2x\). To find \( \frac{dw}{dt} \) using the chain rule, identify the dependent and independent variables. Here, \(w\) is dependent on \(x\) and \(y\), which are dependent on \(t\). Therefore, express \(w\) as a function of \(x, y\), and then \(x\) and \(y\) as functions of \(t\).

Step 2: Apply the Chain Rule

Using the chain rule, \( \frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt} \). Compute the partial derivatives \(\frac{\partial w}{\partial x}\) and \(\frac{\partial w}{\partial y}\).

Step 3: Compute Partial Derivatives

Calculate: \( \frac{\partial w}{\partial x} = \frac{\partial}{\partial x} (x^2y - y^2x) = 2xy - y^2 \)\( \frac{\partial w}{\partial y} = \frac{\partial}{\partial y} (x^2y - y^2x) = x^2 - 2yx \)

Step 4: Differentiate \(x\) and \(y\) with respect to \(t\)

Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):\(x = \cos t \Rightarrow \frac{dx}{dt} = -\sin t\)\(y = \sin t \Rightarrow \frac{dy}{dt} = \cos t\)

Step 5: Substitute and Simplify

Substitute back the derivatives into the chain rule formula and simplify:\[ \frac{dw}{dt} = (2xy - y^2)(-\sin t) + (x^2 - 2yx)(\cos t) \]Plug \(x = \cos t\) and \(y = \sin t\) into this expression:\[ \frac{dw}{dt} = (2\cos t \sin t - \sin^2 t)(-\sin t) + (\cos^2 t - 2\cos t \sin t)(\cos t) \]

Step 6: Final Expression Calculation

Simplify the expression:\[ \frac{dw}{dt} = (-2\cos t \sin^2 t + \sin^3 t) + (\cos^3 t - 2\cos^2 t \sin t) \]Combine like terms to get the final expression.Further simplification will lead you to:\[ \frac{dw}{dt} = \cos t (\cos^2 t - 2 \sin^2 t) + \sin t (\sin^2 t - 2 \cos^2 t) \]

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They help in understanding how a function changes as each variable is varied independently, while the other variables are held constant. Simply put, a partial derivative of a function with respect to one of the variables measures how the function changes as that variable changes slightly, with other variables remaining unchanged. For a function of two variables, such as \( w = x^2y - y^2x \), you can find partial derivatives by treating one variable as a constant and differentiating with respect to the other.- To compute \( \frac{\partial w}{\partial x} \), consider \( y \) as a constant, resulting in \( 2xy - y^2 \).- Conversely, for \( \frac{\partial w}{\partial y} \), treat \( x \) as constant, yielding \( x^2 - 2yx \).Understanding these calculations is crucial for applying the Chain Rule to derive further meaning from a function.

Product and Quotient Rule

In calculus, the product and quotient rules simplify differentiating expressions that involve products or divisions of functions. These rules are particularly useful when functions are composed of two or more functions multiplied or divided together. - The **product rule** is used when you're differentiating a product of two functions, \( u \cdot v \). Its formula is \( (uv)' = u'v + uv' \).- The **quotient rule** applies when differentiating a division of functions, \( \frac{u}{v} \). The formula is \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).In the provided exercise, the product rule is implicit in computing partial derivatives. For \(w = x^2y - y^2x\), identifying each term separately allows you to act as though you're applying the product rule, due to the combinations of \(x\) and \(y\). This understanding aids in the precise differentiation of each term in the product.

Trigonometric Functions

Trigonometric functions such as sine and cosine are vital in calculating derivatives, especially in problems involving parametric equations or those with cyclical properties. In this exercise, \( x = \cos t \) and \( y = \sin t \) represent values that continuously change with \( t \), a common scenario in calculus. - **Sine Function**: Differentiating \( \sin t \) results in \( \cos t \).- **Cosine Function**: Differentiating \( \cos t \) gives \(-\sin t\).These functions are particularly useful when calculating the derivative \( \frac{d w}{d t} \). They allow for transformation of the problem into terms that are more intuitive, such as representing angles or cycles in a real-world context. With these derivatives, the changes in \( x \) and \( y \) with respect to \( t \) can be comprehensively calculated.

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It is a powerful tool used to model and solve problems involving change and motion. The exercise involves key calculus concepts like the chain rule, which provides a method to differentiate composite functions. - **Chain Rule**: This rule is used to differentiate a function based on its dependent variables. It's expressed as: \( \frac{df}{dt} = \frac{df}{du} \cdot \frac{du}{dt} \).In the scenario provided, since \(w\) is a function of both \(x\) and \(y\), each of which are functions of \(t\), the chain rule helps us express the derivative of \(w\) in terms of derivatives with respect to \(x\) and \(y\), ultimately expressing everything in terms of \(t\). This reveals the rate of change of one variable concerning another in a complex, interconnected system, showcasing calculus's versatility and importance in mathematics.