Problem 51 - Calculus Single Variable

Question

Find a line that is tangent to the graph of the given function \(f\) and that is parallel to the line \(y=12 x\). $$ f(x)=3 x^{2}+1 $$

Step-by-Step Solution

Answer

The tangent line is \(y = 12x - 11\).

Step 1: Identify the slope of the tangent line

The problem states that the tangent line should be parallel to the line \(y = 12x\). The slope of this line is 12. Therefore, the slope of the tangent line we are looking for is also 12.

Step 2: Find the derivative of the function

To find the slope of the tangent line to the function \(f(x) = 3x^2 + 1\), we need to calculate its derivative. The derivative \(f'(x)\) of the function \(f(x) = 3x^2 + 1\) is given by \(f'(x) = 6x\).

Step 3: Set the derivative equal to the desired slope

Now, we set \(f'(x) = 12\) because the tangent line's slope must equal the slope of the given parallel line. Solving the equation \(6x = 12\) gives \(x = 2\).

Step 4: Find the point of tangency

Now that we know \(x = 2\), we substitute \(x = 2\) into the function to find the corresponding \(y\) value. \(f(2) = 3(2)^2 + 1 = 12 + 1 = 13\). Therefore, the point of tangency is \((2, 13)\).

Step 5: Write the equation of the tangent line

Using the point-slope form of the equation of a line, \(y - y_1 = m(x - x_1)\), where \((x_1, y_1) = (2, 13)\) and \(m = 12\), we have: \(y - 13 = 12(x - 2)\). Simplifying this, the equation of the tangent line is \(y = 12x - 24 + 13 = 12x - 11\).

Key Concepts

Parallel Lines

Parallel lines are an essential concept in geometry and calculus. They never meet, regardless of how extended they are, because they have equal slopes.
This means that they rise over run at the same rate. For example, if we have a line with the equation \(y = 12x\), the slope is 12, meaning for every unit increase in \(x\), \(y\) increases by 12.
In the context of a tangent line problem, when we say the tangent line must be parallel to a given line, it implies that their slopes need to match.

• The original line: \(y = 12x\)
• The desired tangent line slope: 12

Understanding that parallel lines enforce equal slopes helps when aligning the tangent line with the given line through derivative calculations.

Derivative

The derivative of a function at a given point provides the slope of the tangent line at that exact point.
The fundamental idea is to calculate how the function's output (\(y\)) changes as the input (\(x\)) changes.
For instance, given a function \(f(x) = 3x^2 + 1\), finding the derivative, \(f'(x)\), involves applying basic rules of differentiation.

• The power rule is used to differentiate terms like \(x^2\).
• For a function like \(3x^2\), the derivative becomes \(6x\).

This derived function \(f'(x) = 6x\) allows us to find the slope of the tangent line at any point \(x\) on the curve. Set it equal to your desired slope to find the tangent line's specific point of tangency.

Slope

Slope is a core concept in understanding lines and their behavior. The slope of a line is a measure of its steepness, usually represented as the letter \(m\).
It calculates the ratio of the vertical change to the horizontal change between two points on the line.
In mathematical terms, this is often defined as "rise over run."

• Positive slope: the line ascends as it moves right.
• Zero slope: the line is horizontal.
• Negative slope: the line descends.

In problems that involve finding a tangent line parallel to another line, determining the slope of both the given line and the potential tangent line is crucial. Once you know that these slopes must match, you can use this information in conjunction with the derivative to find the point of tangency.

Point of Tangency

The point of tangency is where a tangent line touches the curve at exactly one point. At this point, the slope of the tangent line matches the derivative of the curve.
Finding this point requires calculating where the derivative equals the slope of the required tangent line.
For example, if we want a tangent line to be parallel to \(y = 12x\), then setting our derived slope function \(f'(x) = 6x\) to 12 determines our \(x\)-coordinate of the tangent point.
Solving \(6x = 12\) yields \(x = 2\).
Next, we substitute \(x = 2\) into the original function \(f(x) = 3x^2 + 1\) to find \(y = 13\).

• This gives the exact point of tangency: \((2, 13)\).

This point is critical as it ensures that the line is tangent to the curve precisely and validates calculations related to the slope and derivative.