Slope Calculation for Curve y = 2x^2

To find the slope of the normal to the curve $y = 2x^2$ at the point where $x = 1$, we first need to determine the slope of the tangent line to the curve at that specific point. The slope of the tangent line is given by the derivative of the function.

Derivative of the Curve

The given curve is defined by the equation: $y = 2x^2$

We find the derivative of $y$ with respect to $x$, denoted as $\frac{dy}{dx}$. This represents the instantaneous rate of change of $y$ with respect to $x$, which is the slope of the tangent line at any point $(x, y)$ on the curve.

Using the power rule for differentiation, which states that $\frac{d}{dx}(ax^n) = anx^{n-1}$: $\frac{dy}{dx} = \frac{d}{dx}(2x^2)$ $\frac{dy}{dx} = 2 \cdot 2x^{(2-1)}$ $\frac{dy}{dx} = 4x$

So, the slope of the tangent line to the curve at any point $x$ is $4x$. Let's denote this as $m_{tangent}$. $m_{tangent} = 4x$

Slope of Tangent at x = 1

We need to find the slope specifically at the point where $x = 1$. We substitute $x = 1$ into the expression for the slope of the tangent: $m_{tangent} \text{ at } x=1 = 4(1)$ $m_{tangent} \text{ at } x=1 = 4$

Thus, the slope of the tangent line to the curve $y = 2x^2$ at $x = 1$ is $4$.

Slope of the Normal Line

The normal line to a curve at a specific point is perpendicular to the tangent line at that same point. The relationship between the slopes of two perpendicular lines (that are not horizontal or vertical) is that their product is $-1$. If $m_{tangent}$ is the slope of the tangent line and $m_{normal}$ is the slope of the normal line, then: $m_{tangent} \times m_{normal} = -1$

To find the slope of the normal line, we rearrange the formula: $m_{normal} = -\frac{1}{m_{tangent}}$

Substituting the value of the slope of the tangent ($m_{tangent} = 4$) at $x=1$: $m_{normal} = -\frac{1}{4}$

Conclusion

The slope of the normal to the curve $y = 2x^2$ at $x = 1$ is $-\frac{1}{4}$. This corresponds to option 4.