Problem 23 - Calculus for Biology and Medicine

Question

(a) Use the formal definition to find the derivative of \(y=\) \(1-x^{3}\) at \(x=2\) (b) Show that the point \((2,-7)\) is on the graph of \(y=1-x^{3}\), and find the equation of the normal line at the point \((2,-7)\). (c) Graph \(y=1-x^{3}\) and the tangent line at the point \((2,-7)\) in the same coordinate system.

Step-by-Step Solution

Answer

(a) The derivative at \( x = 2 \) is \(-12\). (b) The normal line is \( y = \frac{1}{12}x - \frac{43}{6} \). (c) Graphs include the cubic equation, tangent at \( x = 2 \), and the normal line.

Step 1: Understand the Derivative Definition

To find the derivative of a function at a particular point using the formal definition, recall that the derivative of a function \( f(x) \) at a point \( x = a \) is given by \[ f'(a) = \lim_{{h \to 0}} \frac{{f(a + h) - f(a)}}{h}.\] Since \( y = 1 - x^3 \), we have \( f(x) = 1 - x^3 \). Our task is to find \( f'(2) \).

Step 2: Apply the Definition of the Derivative

Substitute \( x = 2 \) into the formula:\[ f'(2) = \lim_{{h \to 0}} \frac{{(1 - (2 + h)^3) - (1 - 2^3)}}{h}.\]Evaluate and simplify the numerator:- \( (2+h)^3 = 8 + 12h + 6h^2 + h^3 \).- Thus, \( 1 - (2+h)^3 = 1 - (8 + 12h + 6h^2 + h^3) = -7 - 12h - 6h^2 - h^3 \).Now, simplify the expression to:\[ \lim_{{h \to 0}} \frac{-7 - 12h - 6h^2 - h^3 + 7}{h} = \lim_{{h \to 0}} \frac{-12h - 6h^2 - h^3}{h}.\]

Step 3: Simplify the Expression

Factor \( h \) out of the numerator:\[ \lim_{{h \to 0}} \frac{h(-12 - 6h - h^2)}{h} = \lim_{{h \to 0}} (-12 - 6h - h^2).\]As \( h \to 0 \), this becomes \(-12\). Therefore, \( f'(2) = -12 \).

Step 4: Verify Point on the Graph

To verify that the point \((2, -7)\) is on the function's graph, substitute \( x = 2 \) into \( y = 1 - x^3 \):\[ y = 1 - (2)^3 = 1 - 8 = -7.\]Thus, the point \((2, -7)\) is indeed on the graph.

Step 5: Find the Equation of the Normal Line

The equation of a line is given by \( y - y_1 = m(x - x_1) \), where \( m \) is the slope. For the normal line, the slope \( m_n \) is the negative reciprocal of the tangent slope \( m_t = -12 \). Therefore, \( m_n = \frac{1}{12} \).Substitute \( (x_1, y_1) = (2, -7) \):\[ y + 7 = \frac{1}{12}(x - 2).\]Simplify to find the equation:\[ y = \frac{1}{12}x - \frac{1}{6} - 7 = \frac{1}{12}x - \frac{43}{6}.\]

Step 6: Graphing the Functions

Graph \( y = 1 - x^3 \), noting that it is a cubic function with a point of interest at \( (2, -7) \). The tangent line at this point has the equation \( y + 7 = -12(x - 2) \) and simplifies to \( y = -12x + 17 \).On the same coordinate system, graph the normal line \( y = \frac{1}{12}x - \frac{43}{6} \), illustrating both line equations at the point \( (2, -7) \). This visually confirms the point and both line interactions.

Key Concepts

Formal Definition of Derivative

When we talk about derivatives, we're dealing with how a function changes at any given point. The formal definition of a derivative provides a way to find the slope of a tangent line to a function at a particular point. Let's break this down simply. Given a function \( f(x) \), the derivative at a certain point \( x = a \) is defined as:

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

The formula involves a limit process where \( h \) approaches zero. Essentially, you're seeing how the function \( f(x) \) behaves as you nudge \( x \) ever so slightly by \( h \).

In our exercise, we apply this definition to the function \( y = 1 - x^3 \) to find its derivative at \( x = 2 \). By substituting into the formula and simplifying, we conclude that the derivative, which is the slope of the tangent line, is \(-12\). This means, at \( x = 2 \), the function is decreasing quite steeply.

Tangent and Normal Lines

Tangent and normal lines give us a deeper understanding of a curve at a point. The tangent line touches the curve at a point without crossing it, much like a gentle whisper of a touch.

In mathematics, the slope of this tangent line to the curve \( y = 1 - x^3 \) at the point \((2, -7)\) is precisely the value of the derivative at that point. From our derivation, this slope is \(-12\).

• The equation for this tangent line is built using the point-slope form: \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is the point on the curve, and \( m \) is the slope.
• For our specific calculation, we have: \[ y + 7 = -12(x - 2) \]
• The complete equation simplifies to: \[ y = -12x + 17 \]

The normal line, however, is perpendicular to the tangent. Its slope is the negative reciprocal of the tangent slope. Hence, for our curve, the normal line's slope is \( \frac{1}{12} \). Accordingly, using the same point, the normal line equation becomes:

• \[ y + 7 = \frac{1}{12}(x - 2) \]
• Or, simplified: \[ y = \frac{1}{12}x - \frac{43}{6} \]