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: Concept Introduction:
The derivative (\(dy/dx\)) at a point on a curve represents the slope of the tangent line at that point. A normal line is perpendicular to the tangent line at the same point. The slopes of perpendicular lines are negative reciprocals.
Step 2: Methodology:
1. Calculate the derivative of the curve's equation: \(m_{\text{tangent}} = \frac{dy}{dx}\).
2. Substitute the given x-value into the derivative to find the tangent slope at that point.
3. Determine the normal slope as the negative reciprocal of the tangent slope: \(m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}\).
Step 3: Detailed Calculation:
Given curve equation: \[ y = 2x^2 \] Compute the derivative of y with respect to x: \[ \frac{dy}{dx} = \frac{d}{dx}(2x^2) = 4x \] This derivative, \(4x\), yields the slope of the tangent at any x-coordinate.
Evaluate the slope at \(x = 1\): \[ m_{\text{tangent}} = \frac{dy}{dx}\bigg|_{x=1} = 4(1) = 4 \] The slope of the tangent at \(x=1\) is 4.
Calculate the slope of the normal line: \[ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{4} \] Step 4: Conclusion:
The slope of the normal to the curve at \(x = 1\) is -1/4.