Problem 25 - Calculus Single Variable

Question

A function \(f\) and a point \(P\) are given. Find the point-slope form of the equation of the tangent line to the graph of \(f\) at \(P\). $$ f(x)=2 x^{2} \quad P=(5,50) $$

Step-by-Step Solution

Answer

The tangent line's equation is \( y - 50 = 20(x - 5) \).

Step 1: Understand the Point-Slope Formula

The point-slope form of a line's equation is given by \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope.

Step 2: Determine the Derivative of the Function

To find the slope \(m\) at point \(P\), we find the derivative of the function \(f(x) = 2x^2\). The derivative \(f'(x)\) describes the slope of the tangent line, calculated as \( f'(x) = \frac{d}{dx}[2x^2]=4x \).

Step 3: Evaluate the Derivative at Point P

Substitute \(x = 5\) into the derivative to find the slope at \(P\). This gives \( f'(5) = 4(5) = 20\).

Step 4: Write the Equation in Point-Slope Form

Substitute the slope \(m = 20\) and point \(P = (5, 50)\) into the point-slope formula: \(y - 50 = 20(x - 5)\).

Step 5: Conclusion: Final Equation of Tangent Line

The point-slope form equation of the tangent line to the graph of \(f(x) = 2x^2\) at the point \(P = (5, 50)\) is \( y - 50 = 20(x - 5) \).

Key Concepts

Tangent Line

A tangent line is a straight line that just "touches" a curve at a particular point, matching the curve's direction at that spot. You can think of it like a line that skims the outer surface of the curve without cutting through it.

To find a tangent line, particularly in exercises like the one where we have the function \(f(x) = 2x^2\) and the point \((5, 50)\), you need to determine the slope at that point. This is because the slope of the tangent line will reflect how steep the curve is at precisely that point.

Once the slope is known, you use the point-slope form of a line's equation to actually write the equation of the tangent line. This is a crucial step in connecting calculus concepts to real-world geometry.

Derivative

The derivative is a fundamental concept in calculus. Think of it like a tool that measures the rate at which something changes.

When we find the derivative of a function, like for \(f(x) = 2x^2\), we're finding a new function \(f'(x)\) that tells us the slope of the original function at any point \(x\).

For polynomials such as \(2x^2\), taking the derivative involves applying simple power rules. Here, \(f'(x) = 4x\). This means that at any point \(x\), the slope of the tangent line can be calculated by simply plugging \(x\) into the derivative.

Slope of a Function

The slope of a function at a particular point is essentially a measure of how steep the graph is at that point.

• If it’s positive, the function is increasing at that point.
• If it’s negative, the function is decreasing.
• If it's zero, the function is level.

To find the slope at a specific point, you use the derivative. By evaluating \(f'(x)\) at a particular \(x\), in this case, 5, you get the slope at that point, which is 20 in our example.

This provides a direct way to understand how the function behaves at different potential points, not just generally, but very specifically.

Calculus Problem-Solving

Solving calculus problems often involves a systematic approach. Understanding the process is key to mastering the subject.

In problems where you need to find a tangent line, start by:

• Identifying the function and the point of tangency.
• Calculating the derivative to determine the slope at the specific point.
• Using the point-slope form of the equation to find the tangent line.

Take, for instance, our problem with \(f(x) = 2x^2\). Once you determine the derivative, plug in the given point, use the slope you find, and write out the equation. This methodical approach helps in breaking down complex problems into manageable parts, making them easier to solve.