Problem 19 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In each of these cases, find the rate of change of \(f(t)\) with respect to \(t\) at the given value of \(t\). a. \(f(t)=t^{3}\left(t^{2}-1\right)\) at \(t=0\) b. \(f(t)=\left(t^{2}-3 t+6\right)^{1 / 2}\) at \(t=1\)

Step-by-Step Solution

Answer

The rate of change of f(t) with respect to t at t = 0 is 0 for part a, and at t = 1 is -\frac{1}{4} for part b.

Step 1: Differentiate function with respect to t for part a

Given the function: \[ f(t) = t^{3}(t^{2} - 1) \]Use the product rule to find its derivative: \[ f'(t) = \frac{d}{dt}[t^{3}] \times (t^{2} - 1) + t^{3} \times \frac{d}{dt}[t^{2} - 1] \]

Step 2: Calculate the derivatives of each part

\[ \frac{d}{dt}[t^{3}] = 3t^{2} \] and \[ \frac{d}{dt}[t^{2} - 1] = 2t \]Substitute these derivatives back into the product rule expression: \[ f'(t) = 3t^{2}(t^{2} - 1) + t^{3}(2t) \]

Step 3: Simplify the derivative

\[ f'(t) = 3t^{2}(t^{2} - 1) + 2t^{4} \]Combine like terms: \[ f'(t) = 3t^{4} - 3t^{2} + 2t^{4} \]Simplify further: \[ f'(t) = 5t^{4} - 3t^{2} \]

Step 4: Evaluate the derivative at t = 0

Substitute t = 0 into the derivative: \[ f'(0) = 5(0)^{4} - 3(0)^{2} = 0 \]

Step 5: Differentiate function with respect to t for part b

Given the function: \[ f(t) = \frac{(t^{2} - 3t + 6)^{1/2}}{2} \]Use the chain rule to find its derivative: \[ f'(t) = \frac{1}{2}(t^{2} - 3t + 6)^{-1/2} \times \frac{d}{dt}[t^{2} - 3t + 6] \]

Step 6: Calculate the inner derivative

\[ \frac{d}{dt}[t^{2} - 3t + 6] = 2t - 3 \]Substitute this back into the chain rule expression: \[ f'(t) = \frac{1}{2}(t^{2} - 3t + 6)^{-1/2} \times (2t - 3) \]This simplifies to: \[ f'(t) = \frac{2t - 3}{2 \times (t^{2} - 3t + 6)^{1/2}} \]

Step 7: Evaluate the derivative at t = 1

Substitute t = 1 into the derivative: \[ f'(1) = \frac{2(1) - 3}{2 \times (1^{2} - 3(1) + 6)^{1/2}} = \frac{2 - 3}{2 \times (1 - 3 + 6)^{1/2}} = \frac{-1}{2 \times \frac{4}{2}} = \frac{-1}{4} \]

Key Concepts

Understanding the Product Rule

The product rule is a fundamental technique in calculus used when differentiating products of two functions. The rule states: if you have two functions, say, \(u(t)\) and \(v(t)\), their derivative is given by:

• \((uv)' = u'v + uv'\)

This means you differentiate one function while keeping the other constant, then switch, and add the results together.

In the exercise, we applied the product rule to the function \(f(t)=t^{3}(t^{2}-1)\). Using the rule, we get:

• \(f'(t) = (t^{3})' (t^{2} - 1) + (t^{3})(t^{2} - 1)'\)

Breaking it down further:

• \((t^{3})' = 3t^{2}\)
• \((t^{2} - 1)' = 2t\)

Putting these back into our equation, we have:

• \(f'(t) = 3t^{2}(t^{2} - 1) + t^{3}(2t)\)

Simplifying this gives us the final derivative, which is \(5t^{4} - 3t^{2}\). The product rule helps break down complex problems into simpler parts.

Mastering the Chain Rule

The chain rule is essential when differentiating composite functions, that is, a function inside another function. The formula for the chain rule is:

• \((f(g(t)))' = f'(g(t)) \times g'(t)\)

In simpler terms, you first find the derivative of the outer function, treating the inner function as a variable, then multiply this by the derivative of the inner function.

For instance, in part b of the exercise, we have \(f(t) = (t^{2} - 3t + 6)^{1/2}\). Applying the chain rule, we get:

• \(f'(t) = \frac{1}{2}(t^{2} - 3t + 6)^{-1/2} \times (2t - 3)\)

Here,

• The outer function is \((x)^{1/2}\), whose derivative is \(\frac{1}{2}x^{-1/2}\)
• The inner function is \(t^{2} - 3t + 6\), whose derivative is \(2t - 3\)

Combining these, we obtain the derivative \(f'(t) = \frac{2t - 3}{2(t^{2} - 3t + 6)^{1/2}}\). The chain rule allows us to navigate through layers of functions effectively.

Basics of Differentiation

Differentiation is a process in calculus that helps us find the rate at which a function is changing at any given point. In other words, it gives us the slope of the function's graph at that point, commonly referred to as the derivative.

Some essential rules of differentiation include:

• The power rule: \((t^n)' = nt^{n-1}\)
• The sum rule: \((u + v)' = u' + v'\)
• The constant rule: \((c)' = 0\) for any constant \(c\)

In the given exercise, for part a, we used the power rule to find the derivatives of \(t^{3}\) and \(t^{2}\). Similarly, for part b, we applied the chain rule along with basic differentiation rules.

Differentiation is about breaking complicated expressions into simpler pieces and then applying these basic rules to find the rate of change. This technique is a cornerstone of calculus and is widely used in various fields such as physics, engineering, and economics.