Question

The slope of the normal to the curve y = \(2x^2\) at x = 1 is:

• A. 4
• B. -4
• C. 1/2
• D. -1/4

Hint:

Remember the relationship between the slopes of perpendicular lines. If a line has slope 'm', any line perpendicular to it will have a slope of '-1/m'. Be careful not to confuse the slope of the tangent with the slope of the normal.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The slope of the tangent to a curve at a point is given by the value of its derivative (\(dy/dx\)) at that point. The normal line is perpendicular to the tangent line at that same point. The slopes of two perpendicular lines are negative reciprocals of each other.
Step 2: Key Formula or Approach:
1. Find the derivative of the curve's equation: \(m_{\text{tangent}} = \frac{dy}{dx}\).
2. Evaluate the derivative at the given x-value to find the slope of the tangent at that point.
3. The slope of the normal is the negative reciprocal of the tangent's slope: \(m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}\).
Step 3: Detailed Explanation:
The equation of the curve is given by: \[ y = 2x^2 \] First, we find the derivative of y with respect to x: \[ \frac{dy}{dx} = \frac{d}{dx}(2x^2) = 2(2x) = 4x \] This expression, \(4x\), gives the slope of the tangent at any point x on the curve.
We need to find the slope at \(x = 1\). So, we substitute \(x = 1\) into the derivative: \[ m_{\text{tangent}} = \frac{dy}{dx}\bigg|_{x=1} = 4(1) = 4 \] The slope of the tangent line at \(x=1\) is 4.
Now, we find the slope of the normal line, which is the negative reciprocal of the tangent's slope. \[ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{4} \] Step 4: Final Answer:
The slope of the normal to the curve at x = 1 is -1/4.