Question

Evaluate the derivative of the following functions at the given point. $$y=t-t^{2} ; t=2$$

Short answer

Answer: The value of the derivative at $$t = 2$$ is -3.

Step-by-step solution

Find the derivative of the function

To find the derivative of the function $$y = t - t^2$$, we'll use the power rule, which states that the derivative of $$t^n$$ is $$nt^{n-1}$$. Applying this rule to both terms, we get: $$\frac{dy}{dt} = \frac{d}{dt}(t - t^2) = \frac{d}{dt}(t) - \frac{d}{dt}(t^2) = 1 - 2t$$ So, the derivative of the function is $$\frac{dy}{dt} = 1 - 2t$$.

Evaluate the derivative at the given point

Now that we have the derivative function, we can evaluate it at the given point $$t = 2$$: $$\frac{dy}{dt}(2) = 1 - 2(2) = 1 - 4 = -3$$ Thus, the derivative of the function $$y = t - t^2$$ at $$t = 2$$ is equal to -3. This means that the slope of the function at the point $$t = 2$$ is -3.

Key concepts

Power Rule

The power rule is one of the most essential tools in calculus for taking derivatives, especially when dealing with polynomial functions. It provides a straightforward way to find the derivative of a monomial (a single term expression in which the variable is raised to a power).
For a function of the form \(t^n\), the power rule states that the derivative is calculated by multiplying the power \(n\) by the variable raised to the power of \(n-1\), expressed as \(nt^{n-1}\).

• For example, if you have \(t^2\), the derivative would be \(2t^{2-1} = 2t\).
• In our original problem, applying the power rule to \(t^2\) gave us \(2t\), while the derivative of \(t\) is simply 1, because \(t^1\) becomes \(1 \cdot t^{1-1} = 1\).

This is a simple yet powerful method that makes taking derivatives of polynomials quick and easy.

Function Evaluation

Function evaluation refers to the process of calculating the value of a function at a particular point. This concept is crucial after deriving a function because it allows us to determine the specific values of derived functions at given input values.
After finding the derivative using the power rule, the next step involves substituting a given value into this derived function. In our example, once we found the derivative \( \frac{dy}{dt} = 1 - 2t\), we needed to evaluate it at \(t = 2\).

• We substituted 2 into the derivative function, giving us \(1 - 2(2) = -3\).
• This process essentially helps us determine how the function changes at that specific point.

Understanding function evaluation is key in applications where knowing the behavior of a function at a certain instant is important, such as physics or engineering.

Slope of a Tangent Line

The concept of a tangent line is vital in understanding derivatives. When we compute the derivative of a function at a particular point, we essentially find the slope of the tangent line at that point.
A tangent line is a straight line that touches a curve at a single point without crossing it. Its slope is a measure of how fast the function is changing at that specific point.

• In the context of our example, the derivative \(-3\) at \(t = 2\) represents this slope.
• This slope tells us that the tangent line to the curve \(y = t - t^2\) at \(t = 2\) is decreasing, since the slope is negative.

Understanding the slope of a tangent line is crucial in various fields such as physics, where it might represent the rate of change of a variable like speed or angle at a particular site.

Calculus

Calculus is a branch of mathematics that deals with rates of change and the accumulation of quantities. It is divided mainly into two areas: differential calculus and integral calculus.
Differential calculus focuses on the concept of a derivative, which measures the rate of change of a particular quantity. This is what we applied in the problem, determining how the function \(y = t - t^2\) changes with respect to \(t\).

• The power rule and other rules in differential calculus make finding derivatives a systematic approach.
• Calculus thus provides us with tools to understand and predict changes in systems, whether they are natural phenomena or man-made processes.

By mastering the basics of calculus, such as derivatives and integrals, one opens up the ability to solve a wide range of practical problems in fields like physics, economics, and engineering.