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 Review:
The derivative \(dy/dx\) at a point represents the slope of the tangent line to the curve at that point. A normal line is perpendicular to the tangent line at the same point. Perpendicular lines have slopes that are negative reciprocals of each other.
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's slope at that point.
3. Determine the normal's slope by taking the negative reciprocal of the tangent's slope: \(m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}\).
Step 3: Detailed Calculation:
Given the curve equation: \[ y = 2x^2 \] First, compute the derivative of y with respect to x: \[ \frac{dy}{dx} = \frac{d}{dx}(2x^2) = 4x \] This expression, \(4x\), yields the tangent slope for any x on the curve.
To find the slope at \(x = 1\), substitute this value into the derivative: \[ m_{\text{tangent}} = \frac{dy}{dx}\bigg|_{x=1} = 4(1) = 4 \] The tangent line at \(x=1\) has a slope of 4.
Next, calculate 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: Conclusion:
The slope of the normal to the curve at \(x = 1\) is -1/4.