Problem 12 - Calculus Early Transcendentals

Question

Use Theorem 7 to find the following derivatives. When feasible, express your answer in terms of the independent variable. $$d z / d t, \text { where } z=\sqrt{r^{2}+s^{2}}, r=\cos 2 t, \text { and } s=\sin 2 t$$

Step-by-Step Solution

Answer

The solution process calculated the derivative of \(z\) with respect to \(t\) and found that \(\frac{dz}{dt} = 0\). This means that the function \(z\) is constant with respect to the variable \(t\), so the expression does not change as \(t\) changes.

Step 1: Find the partial derivatives of \(z\) with respect to \(r\) and \(s\)

First, we need to find the partial derivatives of \(z\) with respect to \(r\) and \(s\). Given that \(z = \sqrt{r^2 + s^2}\), we can write: $$\frac{\partial z}{\partial r} = \frac{1}{2\sqrt{r^2 + s^2}} \cdot 2r = \frac{r}{\sqrt{r^2 + s^2}}$$ and $$\frac{\partial z}{\partial s} = \frac{1}{2\sqrt{r^2 + s^2}} \cdot 2s = \frac{s}{\sqrt{r^2 + s^2}}$$

Step 2: Find the derivatives of \(r\) and \(s\) with respect to \(t\)

Now, we need to find the derivatives of \(r\) and \(s\) with respect to the independent variable \(t\). Given that \(r = \cos 2t\) and \(s = \sin 2t\), we find: $$\frac{dr}{dt} = \frac{d(\cos 2t)}{dt} = -2\sin 2t$$ and $$\frac{ds}{dt} = \frac{d(\sin 2t)}{dt} = 2\cos 2t$$

Step 3: Apply the Chain Rule to find \(\frac{dz}{dt}\)

We now apply the Chain Rule to find the derivative of \(z\) with respect to \(t\): $$\frac{dz}{dt} = \frac{\partial z}{\partial r} \cdot \frac{dr}{dt} + \frac{\partial z}{\partial s} \cdot \frac{ds}{dt}$$ Substituting the values we found in Steps 1 and 2: $$\frac{dz}{dt} = \frac{r}{\sqrt{r^2 + s^2}} \cdot (-2\sin 2t) + \frac{s}{\sqrt{r^2 + s^2}} \cdot (2\cos 2t)$$

Step 4: Substitute \(r\) and \(s\) with their corresponding expressions

We know that \(r = \cos 2t\) and \(s = \sin 2t\). Substituting these values back into the equation we derived in Step 3: $$\frac{dz}{dt} = \frac{\cos 2t}{\sqrt{(\cos 2t)^2 + (\sin 2t)^2}} \cdot (-2\sin 2t) + \frac{\sin 2t}{\sqrt{(\cos 2t)^2 + (\sin 2t)^2}} \cdot (2\cos 2t)$$

Step 5: Simplify the expression

We now simplify the expression: $$\frac{dz}{dt} = -2\sin 2t\cos 2t + 2\cos 2t\sin 2t$$ Upon simplification, we find: $$\frac{dz}{dt} = 0$$ Thus, the derivative of \(z\) with respect to \(t\) is: $$\frac{dz}{dt} = 0$$

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus that allows us to find the derivative of a composite function. In simple terms, it tells us how to differentiate a function that is nested within another function, much like peeling layers of an onion.

For instance, when you have a function like \( z = \sqrt{r^2 + s^2} \), where \( r \) and \( s \) are themselves functions of another variable \( t \), the chain rule becomes particularly essential.

Let's break it down:

• First, you find the partial derivatives of \( z \) with respect to its inner variables \( r \) and \( s \).
• Then, you find the derivatives of \( r \) and \( s \) with respect to the independent variable \( t \).
• Finally, you combine these derivatives using the chain rule, which in its simplest form is: \( \frac{dz}{dt} = \frac{\partial z}{\partial r} \cdot \frac{dr}{dt} + \frac{\partial z}{\partial s} \cdot \frac{ds}{dt} \).

Applying the chain rule correctly ensures you account for how inner functions affect the change in the composite function regarding the variable.

Partial Derivatives

Partial derivatives are calculations that reveal how a multivariable function changes as one of the variables changes, while the other variables are held constant. Think of it as seeing how changing one ingredient affects the flavor of a dish, assuming all the other ingredients remain the same.

For the problem \( z = \sqrt{r^2 + s^2} \), you need to determine how \( z \) changes with small changes in \( r \) and separately with \( s \). This gives rise to what we call partial derivatives:

• The partial derivative with respect to \( r \), noted as \( \frac{\partial z}{\partial r} = \frac{r}{\sqrt{r^2 + s^2}} \), indicates the rate of change of \( z \) with \( r \), holding \( s \) constant.
• Similarly, \( \frac{\partial z}{\partial s} = \frac{s}{\sqrt{r^2 + s^2}} \) shows how \( z \) changes with \( s \), with \( r \) fixed.

Understanding these allows us to piece together how changes in both input variables impact \( z \), especially useful when combining with other rules like the chain rule.

Trigonometric Functions

Trigonometric functions are basic yet essential elements of calculus, governing the relationships within triangles and allowing us to solve problems involving periodic patterns.

In this context, you are given expressions \( r = \cos 2t \) and \( s = \sin 2t \). Here, \( \cos \) and \( \sin \) are trigonometric functions representing the cosine and sine of an angle, respectively. These functions are inherently linked to circular motion and oscillations.

• The derivative of the cosine function, \( \frac{d}{dt}(\cos 2t) = -2\sin 2t \), shows that the rate of change of the cosine function corresponds to the negative sine function, indicating a shift in phase.
• Similarly, the derivative of the sine function, \( \frac{d}{dt}(\sin 2t) = 2\cos 2t \), reflects how the sine function's rate of change results in a cosine function, again demonstrating periodic shifts.

These calculations reflect the cyclical nature of trigonometric functions and their derivatives, tying in beautifully with the synchronization of periodicity and calculus.