Problem 32 - Calculus Early Transcendentals

Question

(a) If \(\mathrm{f}(\mathrm{t})=\mathrm{t}^{2}-\sqrt{\mathrm{t}},\) find \(\mathrm{f}^{\prime}(\mathrm{t}) .\) (b) Check to see that your answer to part (a) is reasonable by comparing the graphs of \(\mathrm{f}\) and \(\mathrm{f}'.\)

Step-by-Step Solution

Answer

(a) \( f'(t) = 2t - \frac{1}{2}t^{-1/2} \); (b) Check graphs for consistency between function shape and slope.

Step 1: Understand the Function

The given function is \( f(t) = t^2 - \sqrt{t} \). It is composed of two parts: \( t^2 \) and \(-\sqrt{t}\). Our aim is to find the derivative \( f'(t) \) of this function.

Step 2: Differentiate the First Term

The first term of the function is \( t^2 \). Using the power rule of differentiation, which states that \( \frac{d}{dt}(t^n) = n \cdot t^{n-1} \), we find the derivative of \( t^2 \): \( \frac{d}{dt}(t^2) = 2t \).

Step 3: Differentiate the Second Term

The second term of the function is \( -\sqrt{t} = -t^{1/2} \). Again using the power rule, we find the derivative: \( \frac{d}{dt}(-t^{1/2}) = -\frac{1}{2}t^{-1/2} \).

Step 4: Combine the Derivatives

Now, we combine the derivatives obtained from the previous steps to find \( f'(t) \):\[ f'(t) = 2t - \frac{1}{2}t^{-1/2} \]

Step 5: Visual Verification

To verify, imagine the graphs of \( f(t) \) and \( f'(t) \). The graph of \( f(t) \) shows the original function, and the derivative \( f'(t) \) represents the slope of \( f(t) \) at any point \( t \). The derivative's value indicates whether \( f(t) \) is increasing or decreasing. By sketching or using graphing technology, you should observe that \( f'(t) \) has critical points where the slope of \( f(t) \) becomes zero or changes sign, which aligns with the function's behavior.

Key Concepts

Power Rule in Derivative Calculus

The power rule is a fundamental tool in calculus for finding derivatives of polynomial functions. It simplifies the differentiation process immensely. When you have a function in the form of a power, such as \( t^n \), the power rule states that the derivative is \( n \cdot t^{n-1} \). For example, in the given function \( f(t) = t^2 - \sqrt{t} \), we apply the power rule to each term separately.

For the first term, \( t^2 \):

• Apply the power rule: \( \frac{d}{dt}(t^2) = 2t \).

For the second term, consider \( -\sqrt{t} \) as \( -t^{1/2} \):

• Apply the power rule: \( \frac{d}{dt}(-t^{1/2}) = -\frac{1}{2}t^{-1/2} \).

This shows how the power rule helps compute derivatives quickly by lowering the power and multiplying by the original exponent.

Graphical Interpretation of Functions and Derivatives

Graphical interpretation allows us to visualize functions and their derivatives, offering deeper insights into their behavior over time. The function \( f(t) = t^2 - \sqrt{t} \) and its derivative \( f'(t) = 2t - \frac{1}{2}t^{-1/2} \) tell different stories when plotted.

For the original function \( f(t) \), the graph represents a curve describing how values of \( f \) change with \( t \). The derivative \( f'(t) \) gives us the slope of this curve at any point \( t \). It essentially measures how steep the function is at various points.

• If \( f'(t) > 0 \): The function is increasing (uphill trend).
• If \( f'(t) < 0 \): The function is decreasing (downhill trend).

By observing the graphs, one can confirm important characteristics, like where the function changes from increasing to decreasing, which is crucial for finding critical points.

Understanding Critical Points

Critical points are significant in calculus because they represent values of \( t \) where the derivatives are zero or undefined. These points indicate potential "turning points" of the function. To find them in \( f(t) = t^2 - \sqrt{t} \), we look where \( f'(t) = 0 \) or is undefined.

From the derivative \( f'(t) = 2t - \frac{1}{2}t^{-1/2} \), solve \( 2t - \frac{1}{2}t^{-1/2} = 0 \) for \( t \). This is a straightforward algebraic process and helps pinpoint the \( t \)-values that flatten the slope.

Here's what to observe at critical points:

• If the derivative changes from positive to negative, \( f(t) \) has a local maximum at that point.
• If it changes from negative to positive, \( f(t) \) has a local minimum.
• If the derivative doesn't change sign, it might be a point of inflexion.

Identifying critical points is key in analyzing functions for peaks, troughs, and intervals of increase or decrease.