Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the gradient \(\nabla f\). $$ f(x, y)=x^{2} y \cos y $$

Step-by-Step Solution

Answer

\( \nabla f = \left( 2x y \cos y, x^{2} \cos y - x^{2} y \sin y \right) \).

Step 1: Understand the Function

The given function is a multivariable function, \(f(x, y)=x^{2} y \cos y\). We are to find its gradient, which involves partial derivatives with respect to each variable.

Step 2: Partial Derivative with respect to x

To find the partial derivative of \(f\) with respect to \(x\), treat \(y\) as a constant:\[\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^{2} y \cos y) = 2x y \cos y\].

Step 3: Partial Derivative with respect to y

To find the partial derivative of \(f\) with respect to \(y\), treat \(x\) as a constant:\[\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^{2} y \cos y)\].Using the product rule, we have \\[= x^{2} \cos y + x^{2} y (-\sin y)\].Simplifying, we get \\[= x^{2} \cos y - x^{2} y \sin y\].

Step 4: Combine Partial Derivatives into Gradient

The gradient \(abla f\) is a vector consisting of the partial derivatives found in Steps 2 and 3:\[abla f = \left( 2x y \cos y, x^{2} \cos y - x^{2} y \sin y \right)\].

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They measure how a function changes as one of the variables changes, while all other variables remain constant. This is similar to taking the derivative in single-variable calculus but expanded to functions with more than one variable.

In the case of the function \(f(x, y)=x^{2} y \, \cos y\), the partial derivative with respect to \(x\) is calculated by treating \(y\) as a constant. The result, \(\frac{\partial f}{\partial x} = 2x y \, \cos y\), shows how the function changes as \(x\) changes, holding \(y\) constant.

Similarly, to find the partial derivative with respect to \(y\), treat \(x\) as constant. The partial derivative \(\frac{\partial f}{\partial y} = x^{2} \, \cos y - x^{2} y \, \sin y\) reflects how \(f\) changes with respect to \(y\) when \(x\) stays the same.

Gradient Vector

The gradient vector is a crucial tool in multivariable calculus, representing the direction and rate of fastest increase of a function. It is denoted by \(abla f\) and comprises partial derivatives with respect to all variables involved in the function.

For the function \(f(x, y)=x^{2} y \, \cos y\), the gradient is found by combining the partial derivatives:

• \(\frac{\partial f}{\partial x} = 2x y \, \cos y\)
• \(\frac{\partial f}{\partial y} = x^{2} \, \cos y - x^{2} y \, \sin y\)

Thus, the gradient vector is:
\[ abla f = \, (2x y \, \cos y, x^{2} \, \cos y - x^{2} y \, \sin y) \].

This vector points in the direction where the function \(f(x, y)\) increases most rapidly and its magnitude gives the rate of this increase.

Product Rule

When dealing with multivariable functions, the product rule from calculus is often used to differentiate terms involving multiplication. The product rule is \[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \].

In context, consider the partial derivative of \(f(x, y)=x^{2} y \, \cos y\) with respect to \(y\). Notice the expression includes products of functions in \(y\).

Using the product rule, where \(u = x^{2}\) and \(v = y \, \cos y\), we calculate:

• \(u = x^{2}\) is a constant relative to \(y\), so \(u' = 0\).
• \(v = y \, \cos y\) requires product rule: derivative of \(y\) times \(\cos y\) plus \(y\) times derivative of \(\cos y\) (which is \(-\sin y\)).

Thus:
\[ \frac{\partial}{\partial y}(x^{2} y \, \cos y) = x^{2} \cos y - x^{2} y \, \sin y \].

Mastering the product rule is essential to tackle complex expressions that are prevalent in multivariable calculus.