find the normal of y=1/x^2 at the point (1,1)

To find the normal line to the curve \( y = \frac{1}{x^2} \) at the point \( (1,1) \), we follow these steps:

### Step 1: Find the derivative (slope of the tangent line)
The derivative of the function \( y = \frac{1}{x^2} \) will give us the slope of the tangent line at any point on the curve.

\[
y = x^{-2}
\]
Differentiate \( y \) with respect to \( x \):

\[
\frac{dy}{dx} = \frac{d}{dx}(x^{-2}) = -2x^{-3}
\]

### Step 2: Evaluate the derivative at the point \( (1,1) \)
Substitute \( x = 1 \) into the derivative to find the slope of the tangent line at \( (1,1) \):

\[
\frac{dy}{dx}\bigg|_{x=1} = -2(1)^{-3} = -2
\]

So, the slope of the tangent line at \( (1,1) \) is \( -2 \).

### Step 3: Find the slope of the normal line
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent is \( -2 \), the slope of the normal line will be:

\[
\text{slope of the normal} = \frac{1}{2}
\]

### Step 4: Write the equation of the normal line
The point-slope form of a line's equation is:

\[
y - y_1 = m(x - x_1)
\]

where \( m \) is the slope and \( (x_1, y_1) \) is the point on the line. Here, the slope is \( \frac{1}{2} \), and the point is \( (1, 1) \). Substituting these values, we get:

\[
y - 1 = \frac{1}{2}(x - 1)
\]

Simplify the equation:

\[
y - 1 = \frac{1}{2}x - \frac{1}{2}
\]
\[
y = \frac{1}{2}x + \frac{1}{2}
\]

Thus, the equation of the normal line to the curve \( y = \frac{1}{x^2} \) at the point \( (1,1) \) is:

\[
y = \frac{1}{2}x + \frac{1}{2}
\]

Would you like a more detailed breakdown, or have any questions?

Here are 5 related questions:
1. How do you find the tangent line of a curve at any given point?
2. What is the geometric meaning of the normal line?
3. How can we interpret the derivative geometrically in this context?
4. Can you find the tangent and normal lines for \( y = x^2 \) at a different point, say \( (2,4) \)?
5. How do you compute higher-order derivatives for more complex curves?

**Tip:** The slope of the normal line is always the negative reciprocal of the slope of the tangent line at that point.

Learn how to find the normal line to the curve y = 1/x^2 at the point (1,1) using calculus concepts. This problem involves calculating the derivative, determining the tangent line's slope, and finding the equation of the normal line. Suitable for students in Grades 10-12.

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