Question

The equation of the tangent to the curve \(y = x^3\) at (1, 1) is

• A. \(3x - y + 2 = 0\)
• B. \(x - 10y - 50 = 0\)
• C. \(3x - y - 2 = 0\)
• D. \(x - 10y + 50 = 0\)

Hint:

To quickly check your answer, ensure two things: 1. The point (1,1) must satisfy the final equation. For option (C), \(3(1) - 1 - 2 = 3-3 = 0\). It works. 2. The slope of the line from the equation must match your calculated slope. For \(3x - y - 2 = 0\), the slope is \(-A/B = -3/(-1) = 3\), which matches.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find the equation of the line that is tangent to the curve \(y=x^3\) at the point (1, 1).

Step 2: Key Formula or Approach:
1. Find the derivative of the function, \(\frac{dy}{dx}\), to get the formula for the slope of the tangent.
2. Evaluate the derivative at the given point to find the numerical slope, \(m\).
3. Use the point-slope form of a linear equation, \(y - y_1 = m(x - x_1)\), to find the equation of the tangent line.

Step 3: Detailed Explanation:
The curve is given by \(y = x^3\).
First, find the derivative with respect to \(x\):
\[ \frac{dy}{dx} = 3x^2 \]
This gives the slope of the tangent at any point \(x\). We need the slope at the point (1, 1). Substitute \(x=1\) into the derivative:
\[ m = 3(1)^2 = 3 \]
The slope of the tangent at (1, 1) is 3.
Now, use the point-slope form with \(m=3\) and the point \((x_1, y_1) = (1, 1)\):
\[ y - y_1 = m(x - x_1) \]
\[ y - 1 = 3(x - 1) \]
Simplify the equation:
\[ y - 1 = 3x - 3 \]
Rearrange it into the general form \(Ax+By+C=0\):
\[ 3x - y - 3 + 1 = 0 \]
\[ 3x - y - 2 = 0 \]

Step 4: Final Answer:
The equation of the tangent to the curve at (1, 1) is \(3x - y - 2 = 0\).