Problem 32
Question
Find the derivative. Assume \(a, b, c, k\) are constants. $$v=a t^{2}+\frac{b}{t^{2}}$$
Step-by-Step Solution
Verified Answer
The derivative is \( \frac{dv}{dt} = 2a t - \frac{2b}{t^3} \).
1Step 1: 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 \( t \).
2Step 2: 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 \( 2a t \).
3Step 3: 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 \( -2b t^{-3} \).
4Step 4: 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 \( 2a t - \frac{2b}{t^3} \).
Key Concepts
Power RuleDifferentiationCalculus Problem-Solving
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:
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: \(-2b t^{-3} \).
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:
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.
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:
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.
Other exercises in this chapter
Problem 32
Find the derivative. Assume that \(a, b, c\), and \(k\) are constants. $$ f(t)=a e^{b t} $$
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Find the equation of the tangent line to \(f(x)=10 e^{-0.2 x}\) at \(x=4\).
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Find the derivative. Assume that \(a, b, c\), and \(k\) are constants. $$ g(\alpha)=e^{\alpha e^{-2 \alpha}} $$
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A fish population is approximated by \(P(t)=10 e^{0.6 t}\), where \(t\) is in months. Calculate and use units to explain what each of the following tells us abo
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