/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Find \(d r / d \theta\). \(\th... [FREE SOLUTION] | 91Ó°ÊÓ

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Find \(d r / d \theta\). \(\theta^{1 / 2}+r^{1 / 2}=1\)

Short Answer

Expert verified
\( \frac{dr}{d\theta} = -r^{1/2} \theta^{-1/2} \).

Step by step solution

01

Understand the Problem

We need to find the derivative \( \frac{dr}{d\theta} \) for the equation \( \theta^{1/2} + r^{1/2} = 1 \). This means we need to determine how \( r \) changes with respect to \( \theta \).
02

Differentiate Both Sides Implicitly

Differentiate both sides of the equation implicitly with respect to \( \theta \). The left side is \( \theta^{1/2} + r^{1/2} \), which gives: \[ \frac{d}{d\theta}(\theta^{1/2}) + \frac{d}{d\theta}(r^{1/2}) = 0 \].
03

Differentiate Each Term

Differentiate \( \theta^{1/2} \) with respect to \( \theta \) to get \( \frac{1}{2} \theta^{-1/2} \). For \( r^{1/2} \), use the chain rule: \( \frac{1}{2} r^{-1/2} \frac{dr}{d\theta} \).
04

Set Up the Equation

Plug these derivatives back into the equation from Step 2: \[ \frac{1}{2} \theta^{-1/2} + \frac{1}{2} r^{-1/2} \frac{dr}{d\theta} = 0 \].
05

Solve for \( \frac{dr}{d\theta} \)

Rearrange the equation to solve for \( \frac{dr}{d\theta} \). Move the first term to the other side: \[ \frac{1}{2} r^{-1/2} \frac{dr}{d\theta} = -\frac{1}{2} \theta^{-1/2} \]. Multiply through by 2 to clear the fractions: \[ r^{-1/2} \frac{dr}{d\theta} = -\theta^{-1/2} \]. Finally, solve for \( \frac{dr}{d\theta} \): \[ \frac{dr}{d\theta} = -r^{1/2} \theta^{-1/2} \].

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Key Concepts

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

Chain Rule
The chain rule is a powerful technique used in calculus to find derivatives of composite functions. When differentiating a composition of two or more functions, the chain rule plays a critical role. In simple terms, it's like peeling layers of an onion, where you differentiate the outer function and then multiply by the derivative of the inner function.

Here is how it generally works:
  • If you have a composite function like \( f(g(x)) \), and you need to differentiate it with respect to \( x \), the chain rule tells us to differentiate \( f \) with respect to \( g \), and multiply by the derivative of \( g \) with respect to \( x \).
  • In mathematical notation, this is expressed as \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \).
When applying the chain rule to the original exercise, we handled the term \( r^{1/2} \) by differentiating it with respect to \( r \), and then multiplying by \( \frac{dr}{d\theta} \) (since \( r \) itself depends on \( \theta \)). This step effectively connects the changing rate of \( r \) with respect to \( \theta \), ensuring accurate computation.
Partial Derivatives
Partial derivatives are used when dealing with functions of multiple variables. Instead of calculating a derivative with respect to just one variable, partial derivatives provide information about how the function changes as one specific variable changes, keeping others constant.

For example:
  • Consider the function \( f(x, y) = x^2 + y^2 \). The partial derivative with respect to \( x \) is obtained by treating \( y \) as a constant, resulting in \( \frac{\partial f}{\partial x} = 2x \).
  • Similarly, the partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = 2y \).
In context with our problem, we have an equation linking \( \theta \) and \( r \), where \( r \) is implicitly a function of \( \theta \). The differentiation activities here involved identifying how changes in \( \theta \) affect \( r \), a concept related to partial differentiation, as it's critical in understanding how each variable contributes to the function’s rate of change.
Derivative Computation
Computing derivatives is a fundamental aspect of calculus, which involves finding the rate at which one quantity changes with respect to another. This computation can be straightforward or involve more complex rules like the product, quotients, or chain rules.

To compute derivatives in our exercise:
  • We first differently evaluated each part of the expression \( \theta^{1/2} + r^{1/2} = 1 \).
  • We used simple differentiation rules for \( \theta^{1/2} \) which gave us \( \frac{1}{2}\theta^{-1/2} \).
  • For \( r^{1/2} \), we applied the chain rule leading to \( \frac{1}{2}r^{-1/2} \cdot \frac{dr}{d\theta} \).
This differentiation was conducted on the concept of implicit differentiation, which stands when an equation defines a function implicitly. By computing derivatives, we essentially unlock the information about how \( r \) changes as \( \theta \) varies, providing the result \( \frac{dr}{d\theta} = -r^{1/2}\theta^{-1/2} \). This answer illuminates the subtle relationships between variables in dynamic systems.

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Most popular questions from this chapter

Find the derivatives of the functions in Exercises \(19-40\) $$ r=\sec \sqrt{\theta} \tan \left(\frac{1}{\theta}\right) $$

Moving along a parabola A particle moves along the parabola \(y=x^{2}\) in the first quadrant in such a way that its \(x\) -coordinate (measured in meters) increases at a steady 10 \(\mathrm{m} / \mathrm{sec.}\) How fast is the angle of inclination \(\theta\) of the line joining the particle to the origin changing when \(x=3 \mathrm{m} ?\)

Graph \(y=-\sin x\) for \(-\pi \leq x \leq 2 \pi .\) On the same screen, graph $$ y=\frac{\cos (x+h)-\cos x}{h} $$ for \(h=1,0.5,0.3,\) and \(0.1 .\) Then, in a new window, try \(h=-1,-0.5,\) and \(-0.3 .\) What happens as \(h \rightarrow 0^{+} ?\) As \(h \rightarrow 0^{-9}\) What phenomenon is being illustrated here?

Radians versus degrees: degree mode derivatives What happens to the derivatives of \(\sin x\) and \(\cos x\) if \(x\) is measured in degrees instead of radians? To find out, take the following steps. a. With your graphing calculator or computer grapher in degree mode, graph $$ f(h)=\frac{\sin h}{h} $$ and estimate \(\lim _{h \rightarrow 0} f(h) .\) Compare your estimate with \(\pi / 180 .\) Is there any reason to believe the limit should be \(\pi / 180 ?\) b. With your grapher still in degree mode, estimate $$ \lim _{h \rightarrow 0} \frac{\cos h-1}{h} $$ c. Now go back to the derivation of the formula for the derivative of sin \(x\) in the text and carry out the steps of the derivation using degree-mode limits. What formula do you obtain for the derivative? d. Work through the derivation of the formula for the derivative of cos \(x\) using degree-mode limits. What formula do you obtain for the derivative? e. The disadvantages of the degree-mode formulas become apparent as you start taking derivatives of higher order. Try it. What are the second and third degree-mode derivatives of sin \(x\) and \(\cos x ?\)

In Exercises \(41-58\) find \(d y / d t\) $$ y=(1+\cos 2 t)^{-4} $$

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