Problem 74 - Calculus for Biology and Medicine

Question

Find the tangent line, in slope-intercept form, of \(y=f(x)\) at the specified point. $$ f(x)=\sqrt{x}\left(x^{3}-1\right), \text { at } x=1 $$

Step-by-Step Solution

Answer

The tangent line at \(x=1\) is \(y = 3x - 3\).

Step 1: Find the derivative of the function

To find the tangent line, we need the derivative of the function, which gives us the slope at any point. Our function is \(f(x) = \sqrt{x}(x^3 - 1)\). We use the product rule for differentiation: if \(u = \sqrt{x}\) and \(v = x^3 - 1\), then the derivative \(f'(x)\) is \((uv)' = u'v + uv'\). Calculate \(u'\) and \(v'\):- \(u = x^{1/2} \Rightarrow u' = \frac{1}{2}x^{-1/2}\)- \(v = x^3 - 1 \Rightarrow v' = 3x^2\)Substitute back into the product rule: \[f'(x) = \frac{1}{2}x^{-1/2}(x^3 - 1) + \sqrt{x}(3x^2)\].

Step 2: Simplify the derivative

Simplify the expression from Step 1:\[f'(x) = \frac{1}{2}x^{1/2}(x^3 - x^{-3/2}) + 3x^{5/2}\].This involves simplifying the terms by combining like terms and factoring where possible.

Step 3: Evaluate the derivative at the given point

Plug \(x = 1\) into \(f'(x)\) to find the slope of the tangent line at \(x=1\):\[f'(1) = \frac{1}{2}(1)^{1/2}(1^3 - 1^{-3/2}) + 3(1)^{5/2}\] = 0 + 3 = 3.

Step 4: Find the y-coordinate of the point on the function

Calculate \(f(1)\) to find the y-coordinate of the point where the tangent line touches the curve:\[f(1) = \sqrt{1}(1^3 - 1) = 0\].Thus, the point of tangency is \((1, 0)\).

Step 5: Write the equation of the tangent line

The tangent line is in slope-intercept form, \(y = mx + c\). Substituting \(m = 3\) from the slope at \(x=1\), and the point \((1, 0)\):The equation is \(y = 3x + c\).Use the point \((1, 0)\) to solve for \(c\): \(0 = 3(1) + c \Rightarrow c = -3\). Thus, the equation is \(y = 3x - 3\).

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus for finding the derivative of a function that is the product of two other functions. Imagine you have a function comprised of two parts, like our function \( f(x) = \sqrt{x}(x^3 - 1) \). This function can be broken into two separate functions: \( u(x) = \sqrt{x} \) and \( v(x) = x^3 - 1 \).
To find the derivative of a product of two functions, the product rule tells us to differentiate each part separately and then combine them as follows:

• First, differentiate \( u(x) \) to get \( u'(x) \).
• Next, differentiate \( v(x) \) to get \( v'(x) \).
• Finally, apply the product rule: \( (uv)' = u'v + uv' \).

Using the product rule, we can differentiate complex products in a systematic way. This technique is essential for cases where simple intervals of differentiation won't work.

Derivative

Derivatives are at the heart of calculus, providing a way to measure how a function changes at any given point. In simpler words, a derivative gives us the slope or steepness of the tangent line to the curve of a function at a particular point.
For example, when we take the derivative of the function \( f(x) = \sqrt{x}(x^3 - 1) \), it helps us understand how the function is behaving at \( x = 1 \). By applying the product rule, we concluded that one of the terms contributing to the derivative is \( \frac{1}{2}x^{-1/2}(x^3 - 1) \) plus the terms arising from differentiating \( x^3 - 1 \).

• The derivative \( f'(x) \) gives us the rate of change of \( f(x) \).
• Using \( f'(1) = 3 \), we know the slope of the tangent line at \( x = 1 \) is 3.

Derivatives are crucial in understanding the dynamic nature of functions and can be applied to motion, growth, and many other real-world phenomena.

Slope-Intercept Form

The slope-intercept form is an equation of a line and is written as \( y = mx + c \) where \( m \) is the slope and \( c \) is the y-intercept. It is a straightforward way to express linear equations and makes it easy to graph a line.
In our problem, after finding the slope of the tangent line using the derivative (which is 3), we substitute it into the slope-intercept equation:

• The slope \( m = 3 \).
• The point of tangency is \( (1, 0) \).

To find the value of \( c \), substitute the point into the equation \( y = 3x + c \) which results in \( c = -3 \). Therefore, the equation of the tangent line is \( y = 3x - 3 \).
This form is very user-friendly for quickly sketching and understanding the basic direction and position of a line on a graph.

Calculus

Calculus is the branch of mathematics that deals with continuous change, much like how geometry deals with shapes and algebra deals with arithmetic operations. It involves two major concepts: differentiation (finding derivatives) and integration (finding integrals).
Differentiation, which was used in this exercise, provides us with the rate of change; it is similar to asking how fast something is changing at a particular instant. To sketch the tangent line to \( y = f(x) \) at a point like \( x = 1 \), calculus empowers us with tools like the derivative and the product rule.

• It offers ways to predict and describe how things grow, shrink, or stay the same.
• By providing insights into the behavior of functions, calculus helps solve practical problems in science, engineering, and economics.

Whether dealing with odd-looking functions or determining precise slopes, calculus reshapes our understanding of mathematical and real-world problems, equipping us with robust tools to analyze and interpret change.