Problem 22 - Calculus for Biology and Medicine

Question

(a) Use the formal definition to find the derivative of \(y=\) \(x^{2}+1\) at \(x=1\) (b) Show that the point \((1,2)\) is on the graph of \(y=x^{2}+1\), and find the equation of the tangent line at the point \((1,+2)\). (c) Graph \(y=x^{2}+1\) and the tangent line at the point \((1,2)\) in the same coordinate system.

Step-by-Step Solution

Answer

(a) The derivative at \(x=1\) is 2. (b) The point (1,2) is on the graph; the tangent line is \(y=2x\). (c) Graph \(y=x^2+1\) and \(y=2x\) with \((1,2)\) as the point of tangency.

Step 1: Use the Definition of Derivative

The derivative of a function at a point \(x = a\) is given by the formula \(f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}\). For the given function \(y = x^2 + 1\), let's find the derivative at \(x=1\).1. Compute \(f(1) = 1^2 + 1 = 2\).2. Compute \(f(1+h) = (1+h)^2 + 1 = 1 + 2h + h^2 + 1 = h^2 + 2h + 2\).3. Substitute into the derivative formula: \[f'(1) = \lim_{h \to 0} \frac{(h^2 + 2h + 2) - 2}{h} = \lim_{h \to 0} \frac{h^2 + 2h}{h}\]4. Simplify the expression: \[= \lim_{h \to 0} \left(h + 2\right)\]5. As \(h\) approaches \(0\), we have \[f'(1) = 2\].

Step 2: Verify Point on Graph and Find Tangent Line Equation

Check if the point \((1, 2)\) lies on \(y = x^2 + 1\):1. Substituting \(x=1\) into the equation, we get \(y = 1^2 + 1 = 2\).2. Hence, the point \((1, 2)\) is indeed on the graph.Next, find the equation of the tangent line:1. The slope of the tangent line at \((1, 2)\) is the derivative at \(x = 1\), which is \(f'(1) = 2\).2. Use the point-slope form of a line, \(y - y_0 = m(x - x_0)\), where \(m\) is the slope and \((x_0, y_0) = (1, 2)\).3. Substitute \(m = 2\), \(x_0 = 1\), and \(y_0 = 2\): \[y - 2 = 2(x - 1)\]4. Simplify to get the equation: \[y = 2x\].

Step 3: Graph the Function and Tangent Line

1. Graph the function \(y = x^2 + 1\), a parabola opening upwards with vertex \((0, 1)\).2. On the same axes, graph the line \(y = 2x\), which crosses the point \((1, 2)\).3. The tangent line meets the parabola at this point, indicating it accurately represents the instantaneous rate of change at \(x = 1\).

Key Concepts

Formal Definition of Derivative

The derivative of a function at a particular point gives us the slope of the tangent line to the function at that point. It's like asking how steep a hill is at a particular place when driving up. The formal definition states that the derivative of a function \( f(x) \) at a point \( x = a \) is:

• \( f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} \)

This limit helps us determine how the function behaves infinitely close to the point. In the exercise, you are given \( f(x) = x^2 + 1 \). To find the derivative at \( x = 1 \), we substitute into the formula.
We calculate \( f(1) = 1^2 + 1 = 2 \) and \( f(1+h) = (1+h)^2 + 1 = h^2 + 2h + 2 \). Substituting these into our formula for the derivative:

• \( f'(1) = \lim_{h \to 0} \frac{h^2 + 2h + 2 - 2}{h} \)

Combining terms gives us \( \frac{h^2 + 2h}{h} \), which simplifies to \( h + 2 \) when the limit is applied. As \( h \) approaches zero, our derivative \( f'(1) \) equals 2. This shows that at \( x = 1 \), the slope of the tangent line is 2.

Tangent Line Equation

Finding the tangent line equation helps us understand how a curve behaves at a precise point. It's the straight line that just touches the curve at a single point without crossing it, showing us the direction of the curve at that exact spot.
To find the equation of the tangent line at a specific point, we use the point-slope form of a line, \( y - y_0 = m(x - x_0) \), where \( m \) is the slope, and \((x_0, y_0) \) is the point of tangency. For our function, we calculated that \( f'(1) = 2 \). So, the slope of the tangent line at \( x=1 \) is 2.
The point (1, 2) is on the curve \( y = x^2 + 1 \). By using the point-slope formula:

• \( y - 2 = 2(x - 1) \)

Simplifying this gives us the tangent line equation:

• \( y = 2x \)

This straight line represents the slope or rate of change of the function at \( x = 1 \), offering a linear approximation of the curve at this point.

Graphing Functions and Tangents

Visualizing the function and its tangent line on the same graph enhances understanding of how the derivative affects curve behavior. The function \( y = x^2 + 1 \) is a parabola that opens upwards, with its vertex at \( (0, 1) \). This graph can be easily visualized by plotting points.
On the same axes, plot the line \( y = 2x \). This line is a visual representation of the slope at the specific point \( (1, 2) \) on the parabola. It crosses the parabola at exactly one point, \((1, 2)\), affirming it as the tangent line.

• The parabola represents the original function.
• The line \( y = 2x \) depicts the rate of change at \( x = 1 \).
• Both graphs share the point \( (1, 2) \), emphasizing the point of tangency.

By looking at the graph, one can see how the tangent line "hugs" the curve at the point, acting as its best linear approximation there. It shows the immediate direction in which the function is moving at that point.