Problem 49 - Calculus an Applied Approach

Question

Find an equation of the line that is tangent to the graph of \(f\) and parallel to the given line. $$ \begin{array}{ll}{\text { Function }} & {\text { Line }} \\\ {f(x)=-\frac{1}{2} x^{3}} & {6 x+y+4=0}\end{array} $$

Step-by-Step Solution

Answer

The equation of the line that is tangent to the graph of the function \(f(x) = -\frac{1}{2}x^3\) and parallel to the given line \(6x + y + 4 = 0\) is \(y = -6x + 6\sqrt{8} + 4\sqrt{2}\).

Step 1: Find the slope of the given line

The equation of a line in slope-intercept form is \(y = mx + b\), where \(m\) represents the slope. The given line is \(6x + y + 4 =0\), which we can reformulate to \(y = -6x - 4\), so the slope of this line, \(m1\), is -6.

Step 2: Calculate the derivative of \(f(x)\)

We need to find the derivative \(f'(x)\) of the function \(f(x) = -\frac{1}{2}x^3\) which gives us the slope of the tangent line at any point x. The derivative of the function \(f(x)\) is \(f'(x) = -\frac{3}{2}x^2\).

Step 3: Find the x-coordinate at which the slope of \(f(x)\) is equal to \(m1\)

Set the derivative \(f'(x)\) equal to \(m1\) and solve for \(x\). This gives us \(-\frac{3}{2}x^2 = -6\). Solving this we get \(x = \sqrt{8}\) or \(x = -\sqrt{8}\). The negative solution does not make sense in this context, so we discard it. Thus, \(x = \sqrt{8}\). It is the x-coordinate where the slope of the tangent line is equal to the slope of the given line.

Step 4: Calculate the y-coordinate of the tangent point

Substitute \(x = \sqrt{8}\) into the equation \(f(x)\) to find the corresponding y-coordinate. Therefore, \(y = f(\sqrt{8}) = -\frac{1}{2}(\sqrt{8})^3 = -4\sqrt{2}\). Thus, the point at which the tangent line touches the function graph is \((\sqrt{8}, -4\sqrt{2})\).

Step 5: Find the equation of the tangent line

We have the slope \(m1 = -6\) and a point \((\sqrt{8},-4\sqrt{2})\) on the line. So, we may use the point-slope form of the line, \(y - y_1 = m1(x - x_1)\) to define our line. Substitute \(m1 = -6\), \(x_1 = \sqrt{8}\), and \(y_1 = -4\sqrt{2}\) to get the equation \(y + 4\sqrt{2} = -6(x - \sqrt{8})\). This simplifies to \(y = -6x + 6\sqrt{8} + 4\sqrt{2}\). So, this is the equation of the line that is tangent to the graph of \(f(x)\) and parallel to the given line.

Key Concepts

Slope of a Line

The slope of a line is a measure of how steep it is. It defines the rate at which the line rises or falls. In a graph, if you imagine moving from one point to another, the slope tells you how much you go up or down for every step you take to the side.
To calculate the slope, we look at the line's equation in the form of \(y = mx + b\), where \(m\) represents the slope. For example, in the given line equation \(6x + y + 4 = 0\), it can be rearranged to \(y = -6x - 4\). Here, the slope \(m\) is -6.

• A positive slope indicates the line rises from left to right.
• A negative slope means it falls from left to right.

Understanding slope is essential for finding tangent lines because the slope of the tangent line matches the slope of the curve at a specific point.

Derivative

A derivative is a powerful concept in calculus that helps you find the rate of change of a function. Think of it as a fancy tool to find the slope of a function at any given point. When we have a function like \(f(x) = -\frac{1}{2}x^3\), its derivative \(f'(x)\) tells us how the function is changing at every point along the curve.
To compute the derivative of \(f(x) = -\frac{1}{2}x^3\), we apply the power rule, resulting in \(f'(x) = -\frac{3}{2}x^2\).

• The derivative gives us the slope of the tangent line at any specific \(x\).
• A zero derivative at a point indicates a horizontal tangent line at that point.

Calculating derivatives is crucial for determining where a line might be tangent to a curve with equal slope.

Parallel Lines

Parallel lines are lines in a plane that never meet. They always stay the same distance apart no matter how far they extend. This concept is straightforward: if two lines have the same slope, they are parallel.
In our context, we found the slope of the given line to be -6. We want our tangent line to have this same slope so it will be parallel to the given line.

• Parallel lines have identical slopes because they don't intersect.
• In equations, only the intercepts differ but the \(x\) coefficients (slope) must match.

Knowing about parallel lines helps us align our tangent line to ensure it runs alongside the given line without crossing it.

Tangent to a Graph

A tangent line to a graph gently brushes a curve at one point, and crucially, it has the same slope as the curve at that point. Imagine leaning a ruler against a round globe, such that it touches it in just one spot and never cuts through.
To find the tangent line mathematically, we equate the derivative of the function to the given slope (from a parallel line). In our exercise, setting \(f'(x) = -6\) helped us figure out that \(x = \sqrt{8}\) was where this magic meeting of graphs happened. We then calculated the y-coordinate by plugging \(x = \sqrt{8}\) into the original function \(f(x)\).

• A tangent line shows the immediate direction of the curve at the point of contact.
• It is useful for approximating function values near the point of tangency.

Understanding tangency ensures that we effectively demonstrate the relationship between lines and curves at specific points.