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Chapter 8: Problem 4

Find the derivative of the function. $$ g(t)=\pi \cos t-\frac{1}{t^{2}} $$

Short Answer

Expert verified
The derivative of the function \( g(t) = \pi \cos t - \frac{1}{t^{2}} \) is \( g'(t) = - \pi \sin t + \frac{2}{t^3} \).

Step by step solution

01

Differentiate the First Function.

Start by differentiating the first function, \(\pi \cos t\). The derivative of a constant times a function is simply the constant times the derivative of the function. The derivative of \(\cos t\) is \(- \sin t\), so the derivative of \(\pi \cos t\) is \(- \pi \sin t\).
02

Differentiate the Second Function.

Next, differentiate the function \(-\frac{1}{t^{2}}\). This can be rewritten as \(-t^{-2}\) for easier computation. The power rule states that the derivative of \(t^n\) is \(n \cdot t^{n-1}\), so the derivative of \(-t^{-2}\) is \(2t^{-3}\) or \(2/t^3\).
03

Combine the Derivatives.

The derivative of the original function g(t), which is the difference of the two functions, 蟺cos(t) and 1/t^2, is the difference of their derivatives. The derivative of the function g(t) is therefore \(- \pi \sin t + \frac{2}{t^3}\).

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus, representing the process of finding the derivative of a function. Think of a derivative as a tool that tells us how a function changes as its input changes. In practical terms, differentiation can help us understand rates of change, like velocity or the slope of a curve.

To differentiate a function, we look for the rate at which the function's output value changes with respect to its input value. When differentiating, we often use rules and formulas to make the process quicker and simpler.
  • The derivative of a function is often denoted using notations like \( f'(x) \) or \( \frac{df}{dx} \).
  • Each function has its own rule for differentiation, and common methods include using differentiation techniques like the power rule and trigonometric differentiation.
  • By following these rules, we can systematically find the derivative of complex functions by breaking them down into simpler components.
Power Rule
The power rule is a convenient method for differentiating functions that involve powers of a variable. It provides a simple formula for finding the derivative of terms that involve variables raised to a power, typically written as \( t^n \).

Here's the formula: if you have a function \( f(t) = t^n \), its derivative \( f'(t) \) is \( n \cdot t^{n-1} \). This means you multiply the power by the variable raised to one lower power.
  • Using the power rule makes it easy to differentiate polynomials and functions like \( t^3 \) or \( t^{-2} \).
  • In our example, \( -\frac{1}{t^2} \) was rewritten as \( -t^{-2} \) before applying the power rule and resulted in the derivative being \( 2t^{-3} \).
  • This technique is handy because it provides a straightforward method for dealing with any function that can be expressed with powers of a variable.
Trigonometric Differentiation
Trigonometric differentiation involves finding the derivative of trigonometric functions like sine, cosine, and tangent. These derivatives have specific rules commonly used in calculus.

For example, the derivatives of the basic trig functions are:
  • \( \frac{d}{dt}(\sin t) = \cos t \)
  • \( \frac{d}{dt}(\cos t) = -\sin t \)
  • \( \frac{d}{dt}(\tan t) = \sec^2 t \)
Applying these rules helps in calculating the rate of change for functions involving trigonometric terms.
  • In the original exercise, the term \( \pi \cos t \) required trigonometric differentiation.
  • The derivative of \( \cos t \) is \( -\sin t \), so multiplying by the constant \( \pi \) gives \( -\pi \sin t \) as the derivative of the term.
  • Using these rules helps in systematically differentiating functions that contain trigonometric expressions and is incredibly useful in calculus problems involving periodic phenomena.

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Most popular questions from this chapter

Sales In Example 9 in Section \(8.4,\) the sales of a seasonal product were approximated by the model $$ F=100,000\left[1+\sin \frac{2 \pi(t-60)}{365}\right], \quad t \geq 0 $$ $$ \begin{array}{l}{\text { where } F \text { was measured in pounds and } t \text { was the time in days, }} \\ {\text { with } t=1 \text { corresponding to January } 1 . \text { The manufacturer }} \\ {\text { of this product wants to set up a manufacturing schedule to }} \\ {\text { produce a uniform amount each day. What should this }} \\ {\text { amount be? (Assume that there are } 200 \text { production days }} \\ {\text { during the year.) }}\end{array} $$

complete the table (using a spreadsheet or a graphing utility set in radian mode) to estimate \(\lim _{x \rightarrow 0} f(x)\). $$ \begin{array}{|c|c|c|c|c|c|c|}\hline x & {-0.1} & {-0.01} & {-0.001} & {0.001} & {0.01} & {0.1} \\ \hline f(x) & {} & {} & {} & {} \\ \hline\end{array} $$ $$ f(x)=\frac{3(1-\cos x)}{x} $$

complete the table (using a spreadsheet or a graphing utility set in radian mode) to estimate \(\lim _{x \rightarrow 0} f(x)\). $$ \begin{array}{|c|c|c|c|c|c|c|}\hline x & {-0.1} & {-0.01} & {-0.001} & {0.001} & {0.01} & {0.1} \\ \hline f(x) & {} & {} & {} & {} \\ \hline\end{array} $$ $$ f(x)=\frac{2 \sin (x / 4)}{x} $$

determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. The period of \(f(x)=5 \cot \left(-\frac{4 x}{3}\right)\) is \(\frac{3 \pi}{2}\)

sketch the graph of the function by hand. Use a graphing utility to verify your sketch. $$ y=\frac{3}{2} \cos \frac{2 x}{3} $$
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