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Chapter 3: Problem 22

Find \(d s / d t\) $$ s=\frac{\sin t}{1-\cos t} $$

Short Answer

Expert verified
The derivative \( \frac{ds}{dt} \) is \( -\frac{1}{1 - \cos t} \).

Step by step solution

01

Identify the quotient rule

We are given a function \( s = \frac{\sin t}{1 - \cos t} \).\ To differentiate a quotient, we'll use the quotient rule: if \( u(t) = \sin t \) and \( v(t) = 1 - \cos t \), then the derivative \( \frac{ds}{dt} \) is given by:\[ \frac{d}{dt}\left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dt} - u \cdot \frac{dv}{dt}}{v^2} \].
02

Differentiate the numerator

Differentiate the numerator \( u(t) = \sin t \). The derivative \( \frac{du}{dt} \) is \( \cos t \).
03

Differentiate the denominator

Differentiate the denominator \( v(t) = 1 - \cos t \). The derivative \( \frac{dv}{dt} \) is \( \sin t \) since the derivative of \(-\cos t\) is \( \sin t \).
04

Apply the quotient rule

Substitute the derivatives and the functions back into the quotient rule formula: \[ \frac{ds}{dt} = \frac{(1 - \cos t) \cdot \cos t - \sin t \cdot \sin t}{(1 - \cos t)^2} \].
05

Simplify the expression

Simplify the expression for \( \frac{ds}{dt} \): \[ \frac{ds}{dt} = \frac{\cos t - \cos^2 t - \sin^2 t}{(1 - \cos t)^2} \].\ Use the identity \( \cos^2 t + \sin^2 t = 1 \) to simplify further: \[ \cos^2 t + \sin^2 t = 1 \implies \sin^2 t = 1 - \cos^2 t \].\ Therefore, \( \cos^2 t + \sin^2 t = 1 \) can help in rewriting the expression as \[ \frac{ds}{dt} = \frac{\cos t - 1}{(1 - \cos t)^2} = \frac{-(1 - \cos t)}{(1 - \cos t)^2} \].\ Finally, simplify to: \[ \frac{ds}{dt} = -\frac{1}{(1 - \cos t)} \].

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Key Concepts

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

Quotient Rule
The quotient rule is a crucial concept in calculus for finding the derivative of a function that is the ratio of two differentiable functions. If you have a function that can be expressed as a fraction, such as \( s = \frac{u(x)}{v(x)} \), you need the quotient rule to differentiate it. Here's the formula that the quotient rule uses:
  • \( \frac{d}{dt}\left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dt} - u \cdot \frac{dv}{dt}}{v^2} \)
To apply it in our specific example, we let \( u(t) = \sin t \) and \( v(t) = 1 - \cos t \). We'll first differentiate both the numerator \( \sin t \) and the denominator \( 1 - \cos t \) with respect to \( t \). Then we plug these derivatives into the quotient rule formula to get our result.
Trigonometry
Trigonometry plays a fundamental role when dealing with derivatives of functions involving sine and cosine. Here, we're working with \( \sin t \) and \( 1 - \cos t \). Understanding the derivatives of these basic trigonometric functions is vital:
  • The derivative of \( \sin t \) is \( \cos t \).
  • The derivative of \( 1 - \cos t \) requires applying the derivative of \( -\cos t \), which is \( \sin t \).
These derivatives are straightforward and relying on trigonometric identities like \( \sin^2 t + \cos^2 t = 1 \) can greatly help in simplifying the expressions further. Using these identities aids in expressing complex trigonometric derivatives in simpler terms, which is particularly useful in the simplification process.
Derivative Simplification
Simplifying the derivative expression is the final step and often involves using trigonometric identities. After applying the quotient rule, we obtained:
  • \( \frac{ds}{dt} = \frac{(1 - \cos t) \cdot \cos t - \sin t \cdot \sin t}{(1 - \cos t)^2} \)
To simplify:
  • Substitute \( \sin^2 t = 1 - \cos^2 t \) from the identity \( \cos^2 t + \sin^2 t = 1 \).
  • This substitution helps in revising \( \cos t - \cos^2 t - \sin^2 t \) to \( \cos t - 1 \).
  • The numerator simplifies to \( -(1 - \cos t) \).
Dividing by \( (1 - \cos t)^2 \), leads us to the final expression: \( -\frac{1}{1 - \cos t} \).
  • Mastering simplification ensures error-free and efficient calculus solutions, and aids in better understanding of function behavior.

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Most popular questions from this chapter

In Exercises \(41-58\) find \(d y / d t\) $$ y=(t \tan t)^{10} $$

The derivative of sin 2\(x\) Graph the function \(y=2 \cos 2 x\) for \(-2 \leq x \leq 3.5 .\) Then, on the same screen, graph $$y=\frac{\sin 2(x+h)-\sin 2 x}{h}$$ for \(h=1.0,0.5,\) and \(0.2 .\) Experiment with other values of \(h\) , in- cluding negative values. What do you see happening as \(h \rightarrow 0 ?\) Explain this behavior.

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 \(\cos x ?\)

Running machinery too fast Suppose that a piston is moving straight up and down and that its position at time \(t\) sec is $$s=A \cos (2 \pi b t)$$ with \(A\) and \(b\) positive. The value of \(A\) is the amplitude of the motion, and \(b\) is the frequency (number of times the piston moves up and down each second). What effect does doubling the frequency have on the piston's velocity, acceleration, and jerk? (Once you find out, you will know why some machinery breaks when you nun it too fast.)

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 \(x=3 \mathrm{m} ?\)
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