Finding the Slope of the Normal to a Curve

To find the slope of the normal to a curve at a given point, we first need to find the slope of the tangent to the curve at that point. The slope of the tangent is obtained by calculating the first derivative of the curve's equation with respect to \(x\).

1. Curve Equation and Point Analysis

• The given curve equation is: \(y = x^2 + 7x\)
• The point at which we need to find the slope of the normal is: \((1, 8)\)

2. Differentiating the Curve to Find Tangent Slope

The slope of the tangent to the curve \(y = f(x)\) at any point is given by the derivative \(\frac{dy}{dx}\). Let's differentiate the given curve equation with respect to \(x\):

Given: \(y = x^2 + 7x\)

Differentiating both sides with respect to \(x\):

\[ \frac{dy}{dx} = \frac{d}{dx}(x^2 + 7x) \] \[ \frac{dy}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}(7x) \]

Using the power rule for differentiation, \(\frac{d}{dx}(x^n) = nx^{n-1}\), and the constant multiple rule, \(\frac{d}{dx}(cf(x)) = c\frac{d}{dx}(f(x))\):

\[ \frac{dy}{dx} = 2x^{2-1} + 7 \cdot 1x^{1-1} \] \[ \frac{dy}{dx} = 2x + 7 \cdot x^0 \] \[ \frac{dy}{dx} = 2x + 7 \cdot 1 \] \[ \frac{dy}{dx} = 2x + 7 \]

This expression, \(2x + 7\), represents the slope of the tangent to the curve \(y = x^2 + 7x\) at any point \((x, y)\).

3. Calculating the Slope of Tangent at (1, 8)

Now, we substitute the \(x\)-coordinate of the given point \((1, 8)\) into the derivative \(\frac{dy}{dx}\) to find the specific slope of the tangent at that point.

Slope of tangent \(m_t = \left(\frac{dy}{dx}\right)_{\text{at } x=1}\)

\[ m_t = 2(1) + 7 \] \[ m_t = 2 + 7 \] \[ m_t = 9 \]

So, the slope of the tangent to the curve \(y = x^2 + 7x\) at the point \((1, 8)\) is \(9\).

4. Determining the Slope of the Normal

The normal to a curve at a point is perpendicular to the tangent at that same point. For two perpendicular lines, the product of their slopes is \(-1\), provided neither slope is zero or undefined.

If \(m_t\) is the slope of the tangent and \(m_n\) is the slope of the normal, then:

\[ m_t \cdot m_n = -1 \]

We found the slope of the tangent \(m_t = 9\). Now we can calculate the slope of the normal \(m_n\):

\[ 9 \cdot m_n = -1 \] \[ m_n = \frac{-1}{9} \]

Therefore, the slope of the normal to the curve \(y = x^2 + 7x\) at the point \((1, 8)\) is \(\frac{-1}{9}\).

5. Summary of Steps for Normal Slope