Problem 4 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Find an equation of the tangent to the graph of \(P\) at the indicated points. Draw the graph \(P\) and the tangent. a. \(P(t)=t^{4} \quad\) at \(\quad(1,1)\) b. \(P(t)=t^{12}\) at \(\quad(1,1)\) c. \(P(t)=t^{1 / 2} \quad\) at \(\quad(4,2)\) d. \(P(t)=\frac{5}{2}\) at e. \(P(t)=\sqrt{1+t}\) at (8,3) f. \(P(t)=\frac{1}{2 t}\) at \(\quad\left(\frac{1}{2}, 1\right)\)

Step-by-Step Solution

Answer

Tangent equations: a. \(y = 4x - 3\), b. \(y = 12x - 11\), c. \(y = \frac{1}{4}x + 1\), e. \(y = \frac{1}{6}x + \frac{10}{3}\), f. \(y = -2x + 2\).

Step 1: Understanding Tangent Line Equation

The equation of a tangent line to the curve at a point can be written as \(y = mx + b\), where \(m\) is the slope of the tangent line and \(b\) is the y-intercept. To find \(m\), take the derivative of the function and evaluate it at the given point.

Step 2: Finding the Derivative

For each part, compute the derivative of the function \(P(t)\): - a. \(P(t) = t^4\) gives \(P'(t) = 4t^3\) - b. \(P(t) = t^{12}\) gives \(P'(t) = 12t^{11}\)- c. \(P(t) = t^{1/2}\) gives \(P'(t) = \frac{1}{2}t^{-1/2}\)- e. \(P(t) = \sqrt{1+t}\) gives \(P'(t) = \frac{1}{2\sqrt{1+t}}\)- f. \(P(t) = \frac{1}{2t}\) gives \(P'(t) = -\frac{1}{2t^2}\).

Step 3: Calculate the Slope at Given Points

Substitute the given points into the derivative to find the slope \(m\): - a. At point (1,1), \(m = 4(1)^3 = 4\)- b. At point (1,1), \(m = 12(1)^{11} = 12\)- c. At point (4,2), \(m = \frac{1}{2}(4)^{-1/2} = \frac{1}{4}\)- e. At point (8,3), \(m = \frac{1}{2\sqrt{1+8}} = \frac{1}{6}\)- f. At point \((\frac{1}{2}, 1)\), \(m = -\frac{1}{2(\frac{1}{2})^2} = -2\).

Step 4: Write Tangent Line Equations

Use the formula \(y = mx + b\): - a. \(y - 1 = 4(x - 1)\) gives \(y = 4x - 3\)- b. \(y - 1 = 12(x - 1)\) gives \(y = 12x - 11\)- c. \(y - 2 = \frac{1}{4}(x - 4)\) gives \(y = \frac{1}{4}x + 1\)- e. \(y - 3 = \frac{1}{6}(x - 8)\) gives \(y = \frac{1}{6}x + \frac{10}{3}\)- f. \(y - 1 = -2(x - \frac{1}{2})\) gives \(y = -2x + 2\).

Step 5: Graphing the Functions and Tangents

Plot each function and its corresponding tangent line on a set of axes: - For each part, first plot \(P(t)\), then sketch the tangent line using the equation derived in the previous step. The tangent line should just touch the graph at the given point and have the appropriate slope.

Key Concepts

Derivatives

The concept of derivatives is central to calculus and helps us understand how a function behaves at any point. When we talk about derivatives, we're referring to the rate at which a function is changing at any given value of its variable. In simple terms, it's like measuring how steep a curve is at a specific point.

To find the derivative of a function like those in the exercise, we usually apply differentiation rules. For polynomial functions, we use the power rule, which states: if you have a function of the form \(at^n\), the derivative is \(nat^{n-1}\).

• For \(P(t) = t^4\), the derivative is \(P'(t) = 4t^3\).
• For \(P(t) = t^{12}\), the derivative is \(P'(t) = 12t^{11}\).

Derivatives are not only for polynomials. You can apply various rules to differentiate roots or rational functions as well. Understanding how to derive a function is the first step towards finding a tangent line equation.

Slope Calculation

Knowing the derivative allows us to calculate the slope of the tangent line at a specific point on the graph of the function. The slope tells us how fast the curve is rising or falling at that exact point.

To find the slope at a given point, you simply evaluate the derivative at that point. For example, if you have the derivative \(P'(t) = 4t^3\) and a point (1,1), you substitute \(t = 1\) into the derivative to find the slope:

• At point (1,1), \(m = 4(1)^3 = 4\).

This slope is crucial because it helps in writing the equation of the tangent line. For each function in the problem, you substitute the value of \(t\) from the given point into the derivative to determine the slope \(m\).
The slope gives us direct information about the steepness of the tangent line at the point.

Tangent Line Equation

The tangent line is a straight line that touches a curve at only one point. Its equation is significant because it provides a very accurate linear approximation of the function near that point.

The general formula for a tangent line equation is \(y = mx + b\), where \(m\) is the slope, and \(b\) is the y-intercept. To write the tangent line equation:

• Use the slope calculated from the derivative at the given point.
• Plug in the coordinate of the point into the formula to solve for \(b\).

For example, using the point-slope formula, \(y - y_1 = m(x - x_1)\), allows you to rearrange and solve for \(y = mx + b\). In part a, with point (1,1) and slope 4, the equation becomes: \(y = 4x - 3\), which tells us the immediate linear behavior of the function near the point (1,1).
A tangent line equation thus provides a direct means to understand function behavior at a specific point.

Graphing Functions

Once we understand the function and its tangent line analytically, it's enlightening to also interpret these graphically. Graphing helps visualize the relationship between the function and its tangent.

To begin, sketch the graph of the function \(P(t)\). These are basic graphs:

• For \(t^4\), the curve rises steeply.
• For \(t^{12}\), the curve is even steeper.

Next, use the tangent line equation you derived to draw the tangent line. This line should just touch the graph at the specific point, illustrating precisely where the function and tangent line share the same slope. For part c, the tangent line at point (4,2) of the curve \(P(t) = t^{1/2}\) will slightly bend and touch the graph, representing gradual change.
Graphing provides a clearer comprehension of how derivatives and tangent lines interact with each other, allowing an intuitive understanding of these calculus concepts.