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.