Question

Evaluate the derivative of the following functions at the given point. $$f(t)=t-t^{2} ; a=2$$

Short answer

Answer: The value of the derivative is $$-3$$.

Step-by-step solution

Finding the derivative of the function

To find the derivative of the function $$f(t) = t - t^2$$ with respect to t, we will apply the power rule. The power rule states that if $$f(t) = t^n$$, where n is a constant, then the derivative $$f'(t) = n \cdot t^{n-1}$$. Apply the power rule to both terms: $$f'(t) = \frac{d}{dt}(t) - \frac{d}{dt}(t^2)$$ $$f'(t) = 1 \cdot t^{1-1} - 2 \cdot t^{2-1}$$ Now simplify the expression: $$f'(t) = 1 - 2t$$

Evaluating the derivative at the given point

Now that we have found the derivative of the function, we can evaluate it at the given point $$t = a = 2$$. Substitute a into the derivative: $$f'(2) = 1 - 2(2)$$ Simplify the expression: $$f'(2) = 1 - 4$$ $$f'(2) = -3$$ So, the derivative of the function $$f(t)$$ at the point $$a = 2$$ is $$-3$$.

Key concepts

Derivative

In calculus, a derivative represents how a function changes as its input changes. It measures the rate at which a function's output values are changing at any given point. Think of it as the slope of the tangent line to the graph of the function at a specified point.

When you find the derivative, you determine how one variable affects another. Suppose you have a function that expresses time along the x-axis and speed along the y-axis. The derivative tells you how quickly speed changes at any given moment in time.

• Derivatives provide critical information about the behavior of functions.
• They help in understanding the rate and direction of change.

In mathematical terms, if you are dealing with the function \(f(t)\), the derivative, denoted by \(f'(t)\), is the function that gives the slope of the curve \(f(t)\) at any given value of \(t\). Whenever you solve problems involving motion, growth, or optimization, derivatives are a key tool.

Power Rule

The Power Rule is one of the most important techniques for finding derivatives in calculus. It makes the process of differentiation remarkably straightforward for polynomial expressions. If you have a function of the form \(f(t) = t^n\), where \(n\) is a constant, the Power Rule states that the derivative is found by bringing down the exponent as a coefficient and reducing the exponent by one.

For example, you apply the Power Rule like this:

• Start with \(t^n\).
• Differentiate: \(f'(t) = n \, t^{n-1}\).

In the exercise \(f(t) = t - t^2\), let's apply the Power Rule separately to each term:

• The derivative of \(t\) is \(1 \times t^{1-1} = 1\).
• The derivative of \(t^2\) is \(2 \times t^{2-1} = 2t\).

So the complete derivative is \(f'(t) = 1 - 2t\). The Power Rule simplifies the often complex process of finding how functions change to a series of simple steps and works seamlessly across terms.

Function Evaluation

Function evaluation involves plugging in specific values for the variable in a function to get a numerical output. Once you have the derivative, the next step commonly involves evaluating it at a particular point to find how the function behaves at that exact location.

Evaluation is a straightforward but crucial step:

• It gives you practical information about the function's rate of change at a specific point.
• In physics, engineering, and economics, evaluating derivatives can indicate maximum and minimum points or steady states of processes.

For the given exercise, after finding \(f'(t) = 1 - 2t\), you evaluate the derivative at \(t = 2\):

• Substitute \(2\) into \(f'(t)\): \(f'(2) = 1 - 2(2)\).
• Simplify to find \(f'(2) = 1 - 4 = -3\).

Thus, the function \(f(t)\)'s rate of change at \(t = 2\) is \(-3\), indicating how fast and in what direction the function is changing at that point.