Problem 18 - 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. \(f(t)=t^{3}-4 t^{2}+5 t \sqrt{t}-5\) at \(t=4\) b. \(f(t)=\frac{2 t^{2}-5}{1-3 t}\) at \(t=-1\)

Step-by-Step Solution

Answer

a. The rate of change at \(t=4\) is 31. b. The rate of change at \(t=-1\) is \(\frac{13}{16}\).

Step 1: Differentiate the given function for part (a)

To find the rate of change of the function with respect to the variable, we need to calculate the derivative. For the function \[f(t) = t^3 - 4t^2 + 5t \sqrt{t} - 5\], we need to differentiate each term individually.The differentiation steps are as follows:\[\frac{d}{dt}(t^3) = 3t^2\]\[\frac{d}{dt}(-4t^2) = -8t\]\[\frac{d}{dt}(5t \sqrt{t}) = \frac{d}{dt}(5t \cdot t^{1/2})= 5\frac{d}{dt}(t^{3/2}) = 5 \cdot \frac{3}{2} t^{1/2} = \frac{15}{2} t^{1/2}\]\[\frac{d}{dt}(-5) = 0\]Combining these derivatives together, we get:\[f'(t) = 3t^2 - 8t + \frac{15}{2} t^{1/2}\]

Step 2: Evaluate the derivative at the given value for part (a)

Substitute the specified value \(t = 4\) into the derivative obtained:\[f'(4) = 3(4^2) - 8(4) + \frac{15}{2} \sqrt{4}\]\[= 3(16) - 32 + \frac{15}{2} \cdot 2\]\[= 48 - 32 + 15\]\[= 31\]

Step 3: Differentiate the given function for part (b)

For the function \(f(t) = \frac{2 t^2 - 5}{1 - 3t}\), we will use the quotient rule:The quotient rule formula is \[\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\]where \(u = 2t^2 - 5\) and \(v = 1 - 3t\), so we need to calculate \(u'\) and \(v'\):\[u' = 4t\]\[v' = -3\]Applying the quotient rule, we get:\[f'(t) = \frac{(4t)(1 - 3t) - (2t^2 - 5)(-3)}{(1 - 3t)^2}\]\[= \frac{4t - 12t^2 + 6t^2 + 15}{(1 - 3t)^2}\]\[= \frac{-6t^2 + 4t + 15}{(1 - 3t)^2}\]

Step 4: Evaluate the derivative at the given value for part (b)

Substitute \(t = -1\) into the derivative obtained:\[f'(-1) = \frac{-6(-1)^2 + 4(-1) + 15}{(1 - 3(-1))^2}\]\[= \frac{-6 + 4 + 15}{(1 + 3)^2}\]\[= \frac{13}{16}\]

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It’s all about finding how a function changes as its input changes.
In other words, it helps us understand the rate of change of functions.
When we differentiate a function, we obtain its derivative. This derivative shows the slope of the function at any given point.
If you have a function like \(f(t) = t^3 - 4t^2 + 5t \sqrt{t} - 5\), differentiating each term reveals how each part contributes to the function's overall rate of change.
For instance, differentiating \(t^3\) gives \(3t^2\), since the power rule tells us to bring down the exponent and subtract one from it.
Similarly, the differentiation of the term \(-4t^2\) gives \(-8t\), and so on for the remaining terms.
Understanding differentiation is key to tackling various problems in calculus.

Quotient Rule

The Quotient Rule is used when you need to differentiate a function that is the ratio of two other functions.
Suppose you have a function given by the ratio \(f(t) = \frac{2t^2 - 5}{1 - 3t}\).
To find its derivative, you'll use the quotient rule which is expressed as: \(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\), where \(u\) and \(v\) are functions of \(t\).
Let's break this down:

• \textbf{Step 1}: Identify \(u\) and \(v\) from your given function. Here \(u = 2t^2 - 5\) and \(v = 1 - 3t\).
• \textbf{Step 2}: Find the derivatives of \(u\) and \(v\): \(u' = 4t\) and \(v' = -3\).
• \textbf{Step 3}: Plug into the Quotient Rule formula: \(f'(t) = \frac{(4t)(1 - 3t) - (2t^2 - 5)(-3)}{(1 - 3t)^2}\).
• \textbf{Step 4}: Simplify the expression to get the final derivative.

By simplifying, we get the derivative ready for evaluation at any specific \(t\)-value.

Evaluating Derivatives

After finding the derivative of a function, the next step is to evaluate it at a given point.
This tells us the exact rate of change of the function at that specific point.
For example, if we have the derivative \(f'(t) = 3t^2 - 8t + \frac{15}{2} t^{1/2}\) and we need to find it at \(t = 4\), we substitute 4 for \(t\) to get:
\(f'(4) = 3 \times 4^2 - 8 \times 4 + \frac{15}{2} \times \sqrt{4}\)
Simplifying, this yields:
\( = 48 - 32 + 15 \
= 31\).
This means the rate of change of the function at \(t = 4\) is 31.
Similarly, for a function given by a quotient, such as \(f(t) = \frac{2 t^2 - 5}{1 - 3t}\), we differentiate it using the quotient rule then substitute \(t = -1\) into the resulting derivative to get the rate of change.
The simplified derivative for the example is: \( f'(t)= \frac{-6t^2 + 4t + 15}{(1 - 3t)^2}\).
Evaluating at \( t = -1 \) gives:
\(f'(-1) = \frac{-6(-1)^2 + 4(-1) + 15}{(1 - 3(-1))^2} = \frac{13}{16}\).
Thus, the rate of change at \( t = -1 \) is \( \frac{13}{16} \).
Evaluating derivatives accurately is crucial for understanding how functions behave at specific points.