Question

Find \(d w / d t\) (a) using the appropriate Chain Rule and (b) by converting \(w\) to a function of \(t\) before differentiating. \(w=x y \cos z, \quad x=t, \quad y=t^{2}, \quad z=\arccos t\)

Short answer

Therefore, the derivative\(dw/dt\), using the chain rule, and by converting \(w\) to a function of \(t\) before differentiating is \(dw/dt = t^2 + t - t^3 / \sqrt{1 - t^2}\)

Step-by-step solution

Part (a): Use the chain rule

The chain rule, applied to \(w(t) = x(t) y(t) cos(z(t))\), provides that \(dw/dt = dx/dt * y * cos(z) + x * dy/dt * cos(z) - x * y * sin(z) * dz/dt\) since the derivative of cos(z) is -sin(z). Plug the given values into the equation, \(dx/dt = 1\), \(dy/dt = 2t\), and \(dz/dt = -1/\sqrt{1 - t^2}\). Substituting these values, we can compute the value of \(dw/dt\).

Part (a): Compute the derivative

After plugged in given values, we got \(dw/dt = t^2 cos(\arccos t) + t cos(\arccos t) - t * t^2 sin(\arccos t) / \sqrt{1 - t^2}\). It simplifies to \(dw/dt = t^2 + t - t^3 / \sqrt{1 - t^2}\)

Part (b): Convert \(w\) to a function of \(t\)

For part b, we convert \(w\) into a function of \(t\) to obtain a single variable function that can be differentiated. From the given equations, we can substitute \(x, y, z\) into the equation for \(w\), yielding \(w = t * t^2 * cos(\arccos t)\) or \(w = t^3 cos(\arccos t)\)

Part (b): Differentiate \(w\)

We then differentiate \(w\), using the chain rule again on \(w = t^3 cos(\arccos t)\) to obtain \(dw/dt = t^2 cos(\arccos t) + t^3 * -sin(\arccos t) * -1 / \sqrt{1 - t^2}\). Simplifying, we get \(dw/dt = t^2 + t - t^3 / \sqrt{1 - t^2}\)

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which something changes. It is the process of calculating the derivative of a function, which represents the function's rate of change at a particular point. In simpler terms, if you have a function that describes how something like speed changes over time, differentiation helps you figure out how quickly the speed is increasing or decreasing at any moment.

For example, if we have a function that describes the position of an object over time, the derivative of that function gives us the velocity, or how fast the object's position is changing with time. Similarly, if we want to know how the area of a circle changes as its radius grows, differentiation can tell us exactly that. In the exercise given, differentiation helps us find the rate at which the variable w changes with respect to t.

Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions of several variables. Unlike single-variable calculus where we deal with functions of one variable (like f(x)), multivariable calculus involves functions with more than one input, such as f(x, y, z). The complexity increases as each variable can affect the function in different ways and at different rates.

The Chain Rule in multivariable calculus comes into play when we want to differentiate a function of several variables that are themselves functions of another variable. Instead of treating each component separately, this rule combines them, acknowledging that the rates of change in the variables are interconnected. For instance, in the exercise given, the function w depends on the variables x, y, and z, each of which is a function of t. Using the Chain Rule, we can express the rate of change of w directly with respect to t, which is essential for understanding how a change in t alone will affect w.

Implicit Differentiation

Implicit differentiation is a technique in calculus used when a function is not given in the standard form y=f(x), but rather in a form where the variables are mixed together, such as an equation in x and y. This technique allows us to find the derivative of y with respect to x without explicitly solving for y as an individual function.

In practice, implicit differentiation involves differentiating both sides of an equation with respect to x, treating y as an implicit function of x. When differentiating terms involving y, we apply the Chain Rule by multiplying by dy/dx (since y is a function of x). This enables us to solve for dy/dx even when the relationship between x and y isn't straightforward. The exercise given doesn't specifically involve implicit differentiation, but the concept is similar to multivariable Chain Rule applications, where we treat each variable's dependency on t implicitly while computing dw/dt.