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Chapter 3: Problem 49

Find \(d y / d t\) $$y=\sin (\cos (2 t-5))$$

Short Answer

Expert verified
\(\frac{dy}{dt} = -2 \cos(\cos(2t-5)) \cdot \sin(2t-5)\).

Step by step solution

01

Identify the Function Composite Layers

The function we have is a composition of functions: an innermost function expressed by the linear expression \(2t-5\), followed by a cosine function \(\cos(z)\), and finally an outer sine function \(\sin(x)\). The goal is to apply the chain rule several times to differentiate.
02

Differentiate the Outermost Function

Start by differentiating \(\sin(u)\), where \(u = \cos(2t-5)\). The derivative of \(\sin(u)\) with respect to \(u\) is \(\cos(u)\). Thus, \(\frac{d}{du}[\sin(u)] = \cos(u)\).
03

Differentiate the Middle Function

Now, differentiate \(\cos(v)\), where \(v = 2t - 5\). The derivative of \(\cos(v)\) with respect to \(v\) is \(-\sin(v)\). Thus, \(\frac{d}{dv}[\cos(v)] = -\sin(v)\).
04

Differentiate the Innermost Function

Differentiate the inner linear expression \(2t - 5\) with respect to \(t\). The derivative is simply \(2\) because it is a linear function. So, \(\frac{d}{dt}[2t-5] = 2\).
05

Apply the Chain Rule

Use the chain rule which states that \(\frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dt}\). Substitute the derivatives found in the previous steps:\[\frac{dy}{dt} = \cos(\cos(2t-5)) \cdot (-\sin(2t-5)) \cdot 2\].
06

Simplify the Expression

Simplify the expression for clarity:\[\frac{dy}{dt} = -2 \cos(\cos(2t-5)) \cdot \sin(2t-5)\]. This is the final expression for the derivative.

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Key Concepts

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

derivative
A derivative is a fundamental concept in calculus. It represents how a function changes as its input changes. In simple terms, it measures the rate of change or the slope of the function at any given point.
Finding the derivative is essential in understanding the behavior of functions, particularly how they evolve over time or react to changes in their variables. When differentiating, we apply specific rules like the power rule, product rule, and chain rule to simplify the process.
A linear expression like \(2t-5\), when differentiated, becomes its coefficient, which is \(2\) here. This straightforward differentiation is due to the constant rate of change inherent in linear functions. Understanding derivatives allows us to model real-world problems with more accuracy and insight.
trigonometric functions
Trigonometric functions such as sine, cosine, and tangent are essential mathematical functions that appear in many areas, from geometry to calculus. They relate the angles of a triangle to the ratios of its sides.
  • Sine Function \( \sin(x) \): It's a periodic function with a wave pattern, crucial in describing oscillations and waves.
  • Cosine Function \( \cos(x) \): Similar to sine, it's also periodic and mirrors the sine wave's behavior but offsets in phase.
In calculus, knowing the derivatives of these functions is key:
  • The derivative of \( \sin(x) \) is \( \cos(x) \).
  • The derivative of \( \cos(x) \) is \( -\sin(x) \).
Understanding these derivatives is necessary when dealing with function compositions involving trigonometric parts, like the function \( \sin(\cos(2t-5)) \). These derived forms enable us to unravel the complex interrelationships of function layers.
function composition
Function composition involves combining two or more functions such that the output of one function becomes the input of another. It's like a chain of operations applied in sequence.
For example, consider the composition \(f(g(h(t)))\) where each function depends on the previous one. To differentiate such compositions, we use the chain rule.
  • This rule states: If you have \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
In our exercise, \( y = \sin(\cos(2t-5)) \) is a composition of three functions:
  • The innermost is a linear function \( 2t-5 \).
  • Middle is the cosine \( \cos(v) \).
  • Outer is the sine \( \sin(u) \).
Each function affects the next, creating a complex but interdependent structure. By following the chain rule, we meticulously differentiate each layer, ultimately simplifying the function's derivative into a usable form. This methodical approach to differentiation highlights the elegance and precision of calculus in handling multi-layered problems.

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Most popular questions from this chapter

Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=\) \(m(x-a)+c\) is a linear function in which \(m\) and \(c\) are constants. If the error \(E(x)=f(x)-g(x)\) were small enough near \(x=a\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions 1\. \(E(a)=0\) 2\. \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\) then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization \(L(x)\) gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \(x-a\)

Find \(y^{\prime \prime}\) $$y=x(2 x+1)^{4}$$

(Continuation of Exercise \(65 .)\) The quotient $$ \frac{f(x+h)-f(x-h)}{2 h} $$ may have a limit as \(h \rightarrow 0\) when \(f\) has no derivative at \(x .\) As a case in point, take \(f(x)=|x|\) and calculate $$ \lim _{h \rightarrow 0} \frac{|0+h|-|0-h|}{2 h} $$ As you will see, the limit exists even though \(f(x)=|x|\) has no derivative at \(x=0 .\) Moral: Before using a centered difference quotient, be sure the derivative exists.

The coordinates of a particle in the metric \(x y\) -plane are differentiable functions of time \(t\) with \(d x / d t=\) \(-1 \mathrm{m} / \mathrm{s}\) and \(d y / d t=-5 \mathrm{m} / \mathrm{s} .\) How fast is the particle's distance from the origin changing as it passes through the point (5,12)\(?\)

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\). Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\) c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying \(|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon\) for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=x^{2 / 3}(x-2), \quad[-2,3], \quad a=2$$
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