Problem 45 - Calculus for Biology and Medicine

Question

Find the tangent line, in standard form, to \(y=f(x)\) at the indicated point. $$ y=2 x^{4}-5 x, \text { at } x=1 $$

Step-by-Step Solution

Answer

The tangent line in standard form is \(3x - y = 6\).

Step 1: Find the derivative of the function

To find the tangent line at a given point, we first need the derivative of the function which gives us the slope of the tangent line. The given function is \(y = 2x^4 - 5x\). The derivative is calculated as follows:\[\frac{dy}{dx} = \frac{d}{dx}(2x^4 - 5x) = 8x^3 - 5.\]

Step 2: Evaluate the derivative at the point

Evaluate the derivative at the point \(x = 1\) to find the slope \(m\) of the tangent line:\[\frac{dy}{dx} \bigg|_{x=1} = 8(1)^3 - 5 = 8 - 5 = 3.\]

Step 3: Calculate the function value at the point

Next, calculate the \(y\)-coordinate of the point on the curve using \(x = 1\):\[y = 2(1)^4 - 5(1) = 2 - 5 = -3.\]So the point is \((1, -3)\).

Step 4: Write the point-slope form of the tangent line

Using the point-slope form of a line equation, \(y - y_1 = m(x - x_1)\), with \(m = 3\) and the point \((1, -3)\), we have:\[y + 3 = 3(x - 1).\]

Step 5: Convert the equation to standard form

Expand and rearrange the point-slope equation to convert it into the standard form, \(Ax + By = C\):\[\begin{align*}y + 3 & = 3x - 3 \y & = 3x - 6 \3x - y & = 6.\end{align*}\]

Key Concepts

Understanding Derivatives

Derivatives are a crucial concept in calculus that help us understand how a function changes. At its core, the derivative of a function at a specific point gives us the slope of the tangent line to the curve at that point. Think of the tangent line as a straight line that just touches the curve at one point, matching the curve's direction exactly at that spot. For a polynomial function like

• The derivative is calculated by differentiating each term of the function with respect to the variable, here, named as "x".
• In the example, the function provided was a polynomial: \(y = 2x^4 - 5x\). We used differentiation rules to get \(\frac{dy}{dx} = 8x^3 - 5\).

To find the slope of the tangent line at a specific point \(x = 1\), we plugged this into the derivative. This gave us a slope \(m = 3\), telling us how steep the tangent line is right at that point.
So, derivatives provide the rate of change and are used to find slopes of functions at specific points.

Point-Slope Form

Once we have the slope from derivatives, we can write an equation for the tangent line using the point-slope form. This form is super handy as it relates a known point on the line and the slope. The equation comes as:

• \(y - y_1 = m(x - x_1)\)
• Here, \(m\) is the slope found using the derivative. \((x_1, y_1)\) is our given point where the tangent touches the curve.

For our example, the slope \(m\) was 3 and the point was (1, -3) which we calculated from the function value at \(x = 1\). Putting it all together:

\(y + 3 = 3(x - 1)\)
This concise equation tells us how the line behaves around the point. It’s simple yet powerful in representing lines with known slope and point.

Standard Form of a Line

Finally, we often want our line in standard form due to its conventional appearance, especially when it comes to comparing or computing easily with other lines. The standard form of a line is written as:

• \(Ax + By = C\)
• "A" and "B" are coefficients of x and y, while "C" is a constant.

To convert from point-slope to standard form, we simply rearrange and simplify the equation:
Starting from \(y + 3 = 3(x - 1)\), we expanded and rearranged:

• First, expand: \(y + 3 = 3x - 3\)
• Then, simplify: \(y = 3x - 6\)
• Rearrange to get: \(3x - y = 6\)

Thus, we have the tangent line in the desired standard form, making it much easier to interpret geometrically and algebraically.