Problem 10 - Calculus

Question

Find \(\Delta y / \Delta x\) and \(d y / d x\) for \(y(x)=1+2 x+3 x^{2}\).

Step-by-Step Solution

Answer

Δy/Δx requires points; dy/dx is 2 + 6x.

Step 1: Finding Δy/Δx

To find the average rate of change \( \Delta y / \Delta x \) of the function \( y(x) = 1 + 2x + 3x^2 \), select two distinct points on the function: \((x_1, y(x_1))\) and \((x_2, y(x_2))\). The formula for the average rate of change is given by:\[ \Delta y / \Delta x = \frac{y(x_2) - y(x_1)}{x_2 - x_1} \]You should plug suitable values into this formula to get a specific value.

Step 2: Finding dy/dx

The instantaneous rate of change or derivative of \( y \) with respect to \( x \) is determined using differentiation. We need to find \( \frac{dy}{dx} \) for the function \( y(x) = 1 + 2x + 3x^2 \).Differentiate each term separately:- The derivative of \( 1 \) is \( 0 \).- The derivative of \( 2x \) is \( 2 \).- The derivative of \( 3x^2 \) is \( 6x \) using the power rule: \( \frac{d}{dx} (ax^n) = nax^{n-1} \).Combine these results:\[ \frac{dy}{dx} = 0 + 2 + 6x = 2 + 6x \]

Key Concepts

Understanding Instantaneous Rate of Change

The instantaneous rate of change is essentially the speed at which a function is changing at any given point. For a function like \( y(x) = 1 + 2x + 3x^2 \), we're interested in understanding how fast \( y \) is changing with respect to \( x \) at a specific moment.
It is similar to the concept of speed in physics. If you think of \( x \) as time and \( y \) as position, then the instantaneous rate of change at a particular point gives you the object's speed at that exact time. For this, we use the derivative \( \frac{dy}{dx} \).
In the solution provided, it is calculated by differentiation as \( 2 + 6x \). This expression allows you to plug in any value of \( x \) to find out how quickly \( y \) is changing at that point.

Grasping Average Rate of Change

The average rate of change is like finding the slope between two points on a graph of a function. It tells us about the overall change between those two points, not just at a single point like instantaneous rate.
To calculate this for \( y(x) = 1 + 2x + 3x^2 \), you choose two different \( x \) values, say \( x_1 \) and \( x_2 \), and find their corresponding \( y \) values. Subtract the two \( y \) values and divide by the difference in the two \( x \) values.
With the formula, \( \Delta y / \Delta x = \frac{y(x_2) - y(x_1)}{x_2 - x_1} \), this method gives you an average view of how the function behaves between two intervals. It's especially useful in predicting trends over longer periods or ranges.

The Differentiation Process

Differentiation is a cornerstone of calculus, a mathematical process used to find the derivative of a function. In simple terms, differentiation is used to determine how a function changes as its inputs change.
In solving for \( \frac{dy}{dx} \) of the function \( y(x) = 1 + 2x + 3x^2 \), each term is differentiated individually.

• The constant term \( 1 \) becomes \( 0 \), as constants don't change.
• The term \( 2x \) turns into \( 2 \).
• The expression \( 3x^2 \) becomes \( 6x \).

These differentiated terms are then summed up to give the final derivative, \( 2 + 6x \). Differentiation allows us to compute the instantaneous rate of change efficiently.

Applying the Power Rule

The power rule is one of the basic techniques used in differentiation, particularly useful for functions that contain polynomial terms. It states that if you have a term like \( ax^n \), the derivative is \( nax^{n-1} \).
When you apply the power rule to \( y(x) = 1 + 2x + 3x^2 \), it simplifies the process of finding the derivative.

• The \( 1 \) being a constant vanishes to \( 0 \).
• The \( 2x \) reduces to \( 2 \times 1x^{1-1} = 2 \).
• For \( 3x^2 \), applying the power rule results in \( 6x \).

By using the power rule, you can quickly and effectively differentiate terms with exponents, making it easier to work out the derivative of polynomial functions.