What is the equation of the normal to the curve:𝑦=2𝑥−1y= 2x−1​ y, equals, square root of, 2, x, minus, 1, end square rootat the point where 𝑥=5x=5x, equals, 5?

1. Break Down the Problem

To find the equation of the normal to the curve at a given point, we need to perform the following steps:

1. Find the derivative (slope of the tangent) of the curve at the point where $x = 5$.
2. Use the slope of the tangent to find the slope of the normal line.
3. Find the coordinates of the point on the curve where $x = 5$.
4. Use the point-slope form of the equation of a line to find the equation of the normal.

2. Relevant Concepts

1. The function of the curve is given as $y = \sqrt{2x - 1}$.
2. The derivative $\frac{dy}{dx}$ gives us the slope of the tangent at any point.
3. The slope of the normal is the negative reciprocal of the slope of the tangent.

3. Analysis and Detail

1. Find the derivative: $y = \sqrt{2x - 1}$ Using the chain rule, the derivative is: $\frac{dy}{dx} = \frac{1}{2\sqrt{2x - 1}} \cdot \frac{d}{dx}(2x - 1) = \frac{1}{2\sqrt{2x - 1}} \cdot 2 = \frac{1}{\sqrt{2x - 1}}$

2. Evaluate the derivative at $x = 5$: $\frac{dy}{dx} \bigg|_{x=5} = \frac{1}{\sqrt{2(5) - 1}} = \frac{1}{\sqrt{10 - 1}} = \frac{1}{\sqrt{9}} = \frac{1}{3}$

3. Find coordinates of the point on the curve: $y = \sqrt{2(5) - 1} = \sqrt{10 - 1} = \sqrt{9} = 3$ So, the point is $(5, 3)$.

4. Find the slope of the normal line: The slope of the normal is the negative reciprocal of $\frac{1}{3}$: $m_{normal} = -3$

5. Using point-slope form to find the equation of the normal: The point-slope form of a line is given by: $y - y_1 = m(x - x_1)$ Substituting the point $(5, 3)$ and the slope $-3$: $y - 3 = -3(x - 5)$ Simplifying this: $y - 3 = -3x + 15$ $y = -3x + 18$

4. Verify and Summarize

We have derived the equation of the normal line at the point $(5, 3)$ to be $y = -3x + 18$. The calculations are confirmed as correct.

Final Answer

The equation of the normal to the curve at the point where $x = 5$ is: $y = -3x + 18$