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Chapter 3: Problem 8

Find the first and second derivatives. $$s=-2 t^{-1}+\frac{4}{t^{2}}$$

Short Answer

Expert verified
The first derivative \(s'\) is \(2t^{-2} - 8t^{-3}\) and the second derivative \(s''\) is \(-4t^{-3} + 24t^{-4}\).

Step by step solution

01

Rewrite the Function

Before taking the derivative, let's rewrite the given function in terms of powers that are easier to differentiate. The given function is \( s = -2 t^{-1} + \frac{4}{t^{2}} \). Rewrite it as \( s = -2t^{-1} + 4t^{-2} \). This makes implementing the power rule more straightforward.
02

Find the First Derivative

To find the first derivative, apply the power rule \(\frac{d}{dt}[t^n] = nt^{n-1}\) to each term. For the first term \(-2t^{-1}\), the derivative is \(-2\cdot(-1)t^{-2} = 2t^{-2}\). For the second term \(4t^{-2}\), the derivative is \(4\cdot(-2)t^{-3} = -8t^{-3}\).Thus, the first derivative \(s'\) is given by:\[s' = 2t^{-2} - 8t^{-3}.\]
03

Find the Second Derivative

Next, we need to differentiate the first derivative \(s' = 2t^{-2} - 8t^{-3}\) to find the second derivative. For the term \(2t^{-2}\), the derivative is \(2\cdot(-2)t^{-3} = -4t^{-3}\). For the term \(-8t^{-3}\), the derivative is \(-8\cdot(-3)t^{-4} = 24t^{-4}\).Thus, the second derivative \(s''\) is:\[s'' = -4t^{-3} + 24t^{-4}.\]

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Key Concepts

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

First Derivative
When working with derivatives in calculus, the first derivative is a way to measure how a function changes as its input changes. It's like asking, "How is the slope of the curve behaving?" In mathematical terms, the first derivative of a function gives us the rate of change or the slope of the tangent line to the function at a particular point. For the exercise we have,
  • The function to differentiate is rewritten as: \(s = -2t^{-1} + 4t^{-2}\)
  • We apply the power rule, which relates to using the formula \(\frac{d}{dt}[t^n] = nt^{n-1}\)
  • The first term \(-2t^{-1}\) after differentiation becomes \(2t^{-2}\)
  • The second term \(4t^{-2}\) becomes \(-8t^{-3}\)
Combining these, the first derivative \(s'\) is \(s' = 2t^{-2} - 8t^{-3}\). This derivative tells us how the original function \(s\) changes with respect to \(t\). Essentially, it shows the function's slope at every point \(t\).
Second Derivative
The second derivative is like taking another snapshot of how the first derivative behaves. When we take the second derivative, we're concerned with how the rate of change itself is changing. This can relate to understanding concavity or acceleration when considering movement. In our exercise, we find:
  • We differentiate \(s' = 2t^{-2} - 8t^{-3}\) again using the power rule
  • For \(2t^{-2}\), applying the power rule gives us \(-4t^{-3}\)
  • For \(-8t^{-3}\), it becomes \(24t^{-4}\)
Thus, the second derivative \(s''\) is \(s'' = -4t^{-3} + 24t^{-4}\). This second derivative tells us about the concavity of the function. If it鈥檚 positive, the function is concave up, and if negative, concave down.
Power Rule
The power rule is an essential tool in calculus for differentiation. It's a simple yet powerful rule to use when dealing with polynomial expressions where variables are raised to powers. According to the power rule, if you have a function \(f(t) = t^n\), its derivative \(f'(t)\) is \(nt^{n-1}\). This rule makes it easy to find derivatives without needing more complex calculus techniques.In our exercise, the power rule simplifies the process:
  • For \(s = -2t^{-1} + 4t^{-2}\), the elements \(-2t^{-1}\) and \(4t^{-2}\) are straightforward to handle using the rule.
  • This rule helped us move from \(-2t^{-1}\) to \(2t^{-2}\), and \(4t^{-2}\) to \(-8t^{-3}\) in the first derivative.
  • Similarly, it guided us in transitioning from \(2t^{-2}\) to \(-4t^{-3}\), and \(-8t^{-3}\) to \(24t^{-4}\) in the second derivative.
Understanding and applying the power rule effectively is crucial, as it lays the foundation for solving more complex calculus problems involving derivatives.

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Most popular questions from this chapter

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Verify that the following pairs of curves meet orthogonally. a. \(x^{2}+y^{2}=4, \quad x^{2}=3 y^{2}\) b. \(x=1-y^{2}, \quad x=\frac{1}{3} y^{2}\)

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. $$f(x)=\frac{x-1}{4 x^{2}+1}, \quad\left[-\frac{3}{4}, 1\right], \quad a=\frac{1}{2}$$
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