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Chapter 3: Problem 32

Find the derivative. Assume \(a, b, c, k\) are constants. $$v=a t^{2}+\frac{b}{t^{2}}$$

Short Answer

Expert verified
The derivative is \( \frac{dv}{dt} = 2a t - \frac{2b}{t^3} \).

Step by step solution

01

Identify the Expression

The given expression is a function of time, denoted by \( v(t) = a t^2 + \frac{b}{t^2} \). To find the derivative, we need to take the derivative of each term with respect to \( t \).
02

Differentiate the First Term

The first term is \( a t^2 \). Using the power rule for differentiation, which is \( \frac{d}{dt}[t^n] = n t^{n-1} \), we find that the derivative of \( a t^2 \) is \( 2a t \).
03

Differentiate the Second Term

The second term is \( \frac{b}{t^2} \), which can be rewritten as \( b t^{-2} \). Using the power rule again, the derivative of \( b t^{-2} \) is \( -2b t^{-3} \).
04

Combine the Derivatives

Now that we have found the derivatives of both terms, we combine them to form the derivative of the entire function. Thus, the derivative \( \frac{dv}{dt} \) is \( 2a t - \frac{2b}{t^3} \).

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Key Concepts

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

Power Rule
The power rule is a fundamental concept in calculus, especially when dealing with polynomials. It makes differentiation easier by providing a quick formula. The rule states: if you have a function of the form \( f(x) = x^n \), its derivative is \( f'(x) = n x^{n-1} \). This rule works for any real number \( n \).

Here's how it applies to the given exercise:
  • For the term \( at^2 \), apply the power rule. The exponent is 2, so bring down the exponent, multiply it by the coefficient \(a\), and subtract one from the exponent: \( 2a t^{2-1} = 2a t \).
  • For \( \frac{b}{t^2} \), which is rewritten as \( b t^{-2} \), the exponent is \(-2\). Thus, using the power rule: \(-2b t^{-3} \).
This rule simplifies the process of finding derivatives, turning potentially complex expressions into routine calculations.
Differentiation
Differentiation is a core operation in calculus that involves calculating the rate at which one quantity changes with respect to another. It's akin to finding the slope of a curve at any point.

In our exercise, differentiation involves:
  • Finding the instantaneous rate of change of the function \(v(t) = a t^2 + \frac{b}{t^2}\) with respect to time \(t\).
  • Applying the power rule to each component to determine its derivative.
The resulting derivative provides a formula that describes how the original function \(v(t)\) behaves as time progresses. Differentiation turns complicated expressions into manageable rates of change, offering insights into their behavior dynamics.
Calculus Problem-Solving
Calculus is about solving problems by analyzing change, and differentiation plays a big role in this process. To tackle any calculus problem, understanding the fundamental rules, like the power rule, is essential.

Here's a structured approach to solving differentiation problems:
  • Identify the Function: Determine the components of the function you need to differentiate. For example, identify both terms in \( v(t)\).
  • Apply Rules: Use the power rule or other relevant rules to find the derivative of each term.
  • Combine Results: Once each component is differentiated, combine them to obtain the complete derivative expression.
By following these steps, you can systematically tackle calculus problems, making even complex functions more approachable and understandable.

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Most popular questions from this chapter

If \(h=f^{n}\), show that $$ \frac{\left(f^{n}\right)^{\prime}}{f^{n}}=n \frac{f^{\prime}}{f} $$

Differentiate the functions in Problems 1-20. Assume that \(A\) and \(B\) are constants. $$ W=4 \cos \left(t^{2}\right) $$

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Show that the relative rate of change of a quotient \(f / g\) is the difference between the relative rates of change of \(f\) and \(g\).

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