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Chapter 12: Problem 12

Use Theorem 12.7 to find the following derivatives. When feasible, express your answer in terms of the independent variable. $$d z / d t, \text { where } z=\sqrt{r^{2}+s^{2}}, r=\cos 2 t, \text { and } s=\sin 2 t$$

Short Answer

Expert verified
Using the Chain Rule and simplifying, we found that the derivative of \(z\) with respect to \(t\) is \(\frac{dz}{dt} = 0\).

Step by step solution

01

Find the derivatives of r and s with respect to t

First, we need to find the derivatives of \(r\) and \(s\) with respect to \(t\). Here's how we do it: $$r = \cos(2t) \Rightarrow \frac{d r}{d t} = -2\sin(2t)$$ $$s = \sin(2t) \Rightarrow \frac{d s}{d t} = 2\cos(2t)$$
02

Apply the Chain Rule

Next, we will apply the Chain Rule to find the derivative of \(z\) with respect to \(t\). The Chain Rule states: $$\frac{dz}{dt} = \frac{dz}{dr}\frac{dr}{dt} + \frac{dz}{ds}\frac{ds}{dt}$$
03

Find the partial derivatives of z

To find \(\frac{dz}{dt}\), we first need to find the partial derivatives of \(z\) with respect to \(r\) and \(s\). Here's how we do it: $$z = \sqrt{r^2 + s^2} \Rightarrow \frac{dz}{dr} = \frac{r}{\sqrt{r^2 + s^2}}$$ $$z = \sqrt{r^2 + s^2} \Rightarrow \frac{dz}{ds} = \frac{s}{\sqrt{r^2 + s^2}}$$
04

Plug in the derivatives into the Chain Rule

Now we can plug in the partial derivatives and the derivatives of \(r\) and \(s\) with respect to \(t\) into the Chain Rule equation: $$\frac{dz}{dt} = \left(\frac{r}{\sqrt{r^2 + s^2}}\right)(-2\sin(2t)) + \left(\frac{s}{\sqrt{r^2 + s^2}}\right)(2\cos(2t))$$
05

Express the result in terms of t (if feasible)

Let's substitute the expressions of \(r\) and \(s\) in terms of \(t\) back into the equation above: $$\frac{dz}{dt} = \left(\frac{\cos(2t)}{\sqrt{\cos^2(2t) + \sin^2(2t)}}\right)(-2\sin(2t)) + \left(\frac{\sin(2t)}{\sqrt{\cos^2(2t) + \sin^2(2t)}}\right)(2\cos(2t))$$ Using \(\cos^2(2t) + \sin^2(2t) = 1\), we simplify and get: $$\frac{dz}{dt} = -2\sin(2t)\cos(2t) + 2\cos(2t)\sin(2t) = 0$$ The derivative of \(z\) with respect to \(t\) is \(\frac{dz}{dt}=0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivatives
In calculus, derivatives are powerful tools that help us understand how functions change. When we talk about finding the derivative of a function, we are looking at how the output of that function changes as the input changes. Imagine you are tracking the speed of a car moving along a road. The derivative, in this case, would tell you the rate at which the car's position is changing at any given point in time. By taking derivatives, we often seek to determine rates of change or the slope of a curve at a specific point.

In our exercise, we find derivatives of two expressions relative to the variable \( t \). These expressions, \( r = \cos(2t) \) and \( s = \sin(2t) \), represent trigonometric functions that depend on \( t \). We calculate their derivatives using basic differentiation rules for trigonometric functions:
  • The derivative of \( \cos(2t) \) is \( -2\sin(2t) \).
  • The derivative of \( \sin(2t) \) is \( 2\cos(2t) \).
These results reveal the rate at which each trigonometric function changes as \( t \) changes.
Trigonometric Functions
Trigonometric functions such as sine and cosine are essential in mathematics, particularly in calculus. These functions describe oscillatory behavior and are a fundamental part of periodic phenomena like sound waves or light. In this exercise, the functions \( r = \cos(2t) \) and \( s = \sin(2t) \) capture the essence of these trigonometric relationships, with the variable \( t \) manipulating their phase.

The process of differentiating trigonometric functions follows specific rules.
  • For the cosine function, \( \cos(kx) \), the derivative is \( -k\sin(kx) \).
  • For the sine function, \( \sin(kx) \), the derivative is \( k\cos(kx) \).
Here, \( k \) stands for any constant multiplying \( x \). In our exercise, \( k = 2 \) since the argument in the trigonometric functions is \( 2t \). This constant affects the wave's frequency, which in terms of derivatives, alters how quickly the function oscillates. Understanding these functions and their derivatives allows us to explore deeper mathematical concepts.
Calculus Theorem
Theorem 12.7 involves the Chain Rule, a fundamental theorem in calculus used when differentiating composite functions. The Chain Rule provides a systematic way to find the derivative of a function composed of other functions by breaking it into manageable parts.

In the given exercise, we use the Chain Rule to find the derivative of \( z = \sqrt{r^2 + s^2} \) with respect to \( t \). This involves:
  • Identifying \( z \) as a composite function in terms of \( r \) and \( s \).
  • Computing partial derivatives of \( z \) concerning \( r \) and \( s \).
  • Using the Chain Rule formula: \( \frac{dz}{dt} = \frac{dz}{dr}\frac{dr}{dt} + \frac{dz}{ds}\frac{ds}{dt} \).
Plugging in the values from earlier calculations gives us the derivative of \( z \) in terms of \( t \). After applying trigonometric identities such as \( \cos^2(x) + \sin^2(x) = 1 \), the expression simplifies, showing robust properties of trigonometric functions that lead to a derivative of zero. This represents a situation where \( z \) does not change with \( t \), showcasing the harmonious balance between the trigonometric components.

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