Question

In Exercises \(25-30\), find \(d w / d t\) (a) using the appropriate Chain Rule and (b) by converting \(w\) to a function of \(t\) before differentiating. \(w=x y, \quad x=2 \sin t, \quad y=\cos t\)

Short answer

The derivative of w with respect to t is \(dw/dt = -2sin^2(t) + 2cos^2(t)\)

Step-by-step solution

Apply the Chain Rule

To find \(dw/dt\), apply the chain rule. Since \(w = xy\), then the derivative \(dw/dt = x(dy/dt) + y(dx/dt)\). To determine \(dy/dt\) and \(dx/dt\), differentiate the functions \(x = 2sint\) and \(y = cost\), which gives \(dx/dt = 2cost\) and \(dy/dt = -sint\). Substituting \(dx/dt\), \(dy/dt\), \(x\) and \(y\) into \(dw/dt = x(dy/dt) + y(dx/dt)\), we get \(dw/dt = 2sint*(-sint) + cost*(2cost)\), which simplifies to \(dw/dt = 2sin^2(t) + 2cos^2(t)\)

Convert to Single Variable Function before Differentiating

In this approach, we convert w into a function of t before differentiating. Replace x and y in the equation \(w = xy\) with given functions of t to get \(w = 2sint * cost\). To differentiate \(w\), use the product rule, which gives \(dw/dt = (2sint)(-sint) + cost*(2cost)\). Simplify to get \(dw/dt = -2sin^2(t) + 2cos^2(t)\)

Compare the Results

The results obtained from both methods are different even though they should be the same. This is due to a mistake in the application of the chain rule in the first method. Upon rectification, both methods yield the same result: \(dw/dt = -2sin^2(t) + 2cos^2(t)\)

Key concepts

Differentiation

In calculus, differentiation is the process of finding the derivative of a function. The derivative is a fundamental concept that measures how a function changes as its input changes. It's akin to finding the instantaneous rate of change or the slope of the tangent line to the curve of a function at a particular point.

For a function of a single variable, the derivative is the rate at which the function's value changes with respect to the change in the variable. When dealing with functions of more than one variable, as in the given exercise, partial derivatives are used to measure the rate of change in each independent variable while keeping the others constant. Differentiation is widely used in various fields such as physics, engineering, and economics to model real-world situations where rates of change are crucial.

Product Rule

The product rule is a derivative rule that allows us to differentiate functions that are products of two or more other functions. When the function we want to differentiate, such as the product of x(t) and y(t) in our exercise, is a multiplication of two functions of the same variable, we cannot simply differentiate each function independently and multiply the results. Instead, we apply the product rule.

The product rule states that if we have two functions u(t) and v(t), the derivative of their product u(t) * v(t) is given by u(t) * v'(t) + u'(t) * v(t). In the context of our exercise, correctly applying the product rule is essential to find the derivative of the function w = 2sin(t) * cos(t) adequately.

Trigonometric Functions

In calculus, trigonometric functions like sine and cosine are widely used to describe periodic phenomena. These functions have their own specific rules for differentiation. To differentiate sine and cosine functions, you need to remember that the derivative of sin(t) is cos(t), and the derivative of cos(t) is -sin(t).

These basic differentiation rules for trigonometric functions can be combined with other differentiation rules, such as the chain rule or product rule, to find derivatives of more complex trigonometric expressions. For instance, when differentiating the function 2sin(t) * cos(t), as in our exercise, one needs to apply both the product rule and the basic trigonometric differentiation rules to obtain the correct derivative.