Problem 48 - Calculus for Biology and Medicine

Question

Find the tangent line, in standard form, to \(y=f(x)\) at the indicated point. $$ y=3 \pi x^{5}-\frac{\pi}{2} x^{3}, \text { at } x=-1 $$

Step-by-Step Solution

Answer

The tangent line at \( x = -1 \) is \( 27\pi x - 2y = 22\pi \).

Step 1: Find the derivative

To find the tangent line, we'll start by finding the derivative of the function, which is the slope of the tangent line. Given the function \[ y = 3\pi x^5 - \frac{\pi}{2} x^3 \] we differentiate term-by-term:\[ y' = \frac{d}{dx}(3\pi x^5) - \frac{d}{dx}(\frac{\pi}{2} x^3) \]Using the power rule, we get:\[ y' = 15\pi x^4 - \frac{3\pi}{2} x^2 \].

Step 2: Calculate the slope at the given point

Next, we need to find the slope of the tangent line at the specific point where \( x = -1 \). Substitute \( x = -1 \) into the derivative to calculate the slope of the tangent line:\[ y'(-1) = 15\pi (-1)^4 - \frac{3\pi}{2} (-1)^2 \]. Calculating this gives:\[ y'(-1) = 15\pi - \frac{3\pi}{2} = \frac{30\pi}{2} - \frac{3\pi}{2} = \frac{27\pi}{2} \]. The slope of the tangent line is \( \frac{27\pi}{2} \).

Step 3: Find the point on the curve

To find the point on the curve where the tangent meets, substitute \( x = -1 \) back into the original function:\[ y = 3\pi (-1)^5 - \frac{\pi}{2}(-1)^3 = -3\pi + \frac{\pi}{2} \].This simplifies to:\[ y = -3\pi + \frac{\pi}{2} = -\frac{6\pi}{2} + \frac{\pi}{2} = -\frac{5\pi}{2} \].So, the point on the curve is \( (-1, -\frac{5\pi}{2}) \).

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

Using the point-slope form of a line \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \((x_1, y_1)\) is the point on the line. Substituting the values we found, \( m = \frac{27\pi}{2} \), \( x_1 = -1 \), \( y_1 = -\frac{5\pi}{2} \), the equation is:\[ y + \frac{5\pi}{2} = \frac{27\pi}{2}(x + 1) \].

Step 5: Convert to standard form

The standard form of a line is \( Ax + By = C \). Start by simplifying the point-slope form:\[ y + \frac{5\pi}{2} = \frac{27\pi}{2}x + \frac{27\pi}{2} \].Rearrange it to:\[ 0 = \frac{27\pi}{2}x - y + \frac{27\pi}{2} - \frac{5\pi}{2} \].Simplify to get:\[ \frac{27\pi}{2}x - y = \frac{22\pi}{2} \].Multiply through by 2 to clear the fraction:\[ 27\pi x - 2y = 22\pi \].So, the standard form of the tangent line is \( 27\pi x - 2y = 22\pi \).

Key Concepts

Tangent line

In calculus, a tangent line is a straight line that just "touches" a curve at a specific point. This means it does not cross the curve at that point, but rather has the same slope as the curve there. If you imagine a circle with a radius, the tangent would be a line that lightly brushes one side of the circle without digging in. Tangent lines are important because they can help us understand how a curve behaves at a particular spot. They capture the instantaneous rate of change of a function, giving us a snapshot of its behavior at a single point. When solving problems involving tangent lines, like those in calculus exercises, you often have to find the slope and the point of tangency to construct the equation of the tangent line.

Derivative

A derivative is a fundamental concept in calculus. It's essentially a tool to understand how a function is changing at any given point. Think of it as the "rate of change" of a function, similar to how the speedometer in a car shows how fast your speed is changing as you press on the gas pedal.
In a mathematical context, the derivative of a function is a new function that gives us the slope of the original function at any point. Finding the derivative requires specific rules, such as the power rule, which makes the process systematic.
In the given problem, by finding the derivative of the function, you are determining the slope of the tangent line at any value of \(x\). This slope is crucial because it acts as the foundation to find the equation of the tangent line.

Power rule

The power rule is an essential shortcut in calculus for finding derivatives, especially when dealing with polynomial functions. It simplifies the process of differentiation, allowing you to quickly find the derivative without getting bogged down in complex calculations.
Here's how it works: If you have a term in the form of \(x^n\), where \(n\) is a real number, the derivative of that term is \(nx^{n-1}\). You simply bring the exponent \(n\) down as a coefficient and subtract 1 from the original exponent.

• For example, the derivative of \(x^2\) is \(2x^{2-1} = 2x\).
• This rule makes differentiating terms like \(3\pi x^5\) and \(-\frac{\pi}{2}x^3\) straightforward, resulting in \(15\pi x^4\) and \(-\frac{3\pi}{2}x^2\), respectively.

The power rule speeds up finding derivatives, making it easier to determine slopes of tangent lines.

Standard form of a line

The standard form of a line is a way of expressing the equation of a line. It is written as \(Ax + By = C\), where \(A\), \(B\), and \(C\) are integers, and \(A\) should ideally be a positive number.
This form helps in various mathematical analyses, such as determining intercepts and making system of equations easier to solve. When you are asked to convert a line equation from another form, like point-slope form, into standard form, you usually rearrange the terms to match \(Ax + By = C\).
In our example, the point-slope form \(y + \frac{5\pi}{2} = \frac{27\pi}{2}(x + 1)\) is rearranged into the standard form \(27\pi x - 2y = 22\pi\). The standard form is particularly useful because it allows you to easily identify whether two lines are parallel or perpendicular by comparing their coefficients.