If x = 1-t^2/1+t^2 and y=2t/1+t^2 then find dy/dx

To find \( \frac{dy}{dx} \), given that \( x = \frac{1 - t^2}{1 + t^2} \) and \( y = \frac{2t}{1 + t^2} \), we need to find \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) and then use the chain rule:

\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}
\]

### Step 1: Find \( \frac{dx}{dt} \)

Given:
\[
x = \frac{1 - t^2}{1 + t^2}
\]

We use the quotient rule for differentiation:
\[
\frac{dx}{dt} = \frac{(1 + t^2) \cdot \frac{d}{dt}(1 - t^2) - (1 - t^2) \cdot \frac{d}{dt}(1 + t^2)}{(1 + t^2)^2}
\]

Calculate the derivatives of the numerator terms:
\[
\frac{d}{dt}(1 - t^2) = -2t \quad \text{and} \quad \frac{d}{dt}(1 + t^2) = 2t
\]

Substitute these values:
\[
\frac{dx}{dt} = \frac{(1 + t^2)(-2t) - (1 - t^2)(2t)}{(1 + t^2)^2}
\]

Simplify:
\[
= \frac{-2t - 2t^3 - 2t + 2t^3}{(1 + t^2)^2} = \frac{-4t}{(1 + t^2)^2}
\]

So,
\[
\frac{dx}{dt} = \frac{-4t}{(1 + t^2)^2}
\]

### Step 2: Find \( \frac{dy}{dt} \)

Given:
\[
y = \frac{2t}{1 + t^2}
\]

Using the quotient rule:
\[
\frac{dy}{dt} = \frac{(1 + t^2) \cdot \frac{d}{dt}(2t) - (2t) \cdot \frac{d}{dt}(1 + t^2)}{(1 + t^2)^2}
\]

Calculate the derivatives:
\[
\frac{d}{dt}(2t) = 2 \quad \text{and} \quad \frac{d}{dt}(1 + t^2) = 2t
\]

Substitute these values:
\[
\frac{dy}{dt} = \frac{(1 + t^2)(2) - (2t)(2t)}{(1 + t^2)^2}
\]

Simplify:
\[
= \frac{2 + 2t^2 - 4t^2}{(1 + t^2)^2} = \frac{2 - 2t^2}{(1 + t^2)^2}
\]

So,
\[
\frac{dy}{dt} = \frac{2(1 - t^2)}{(1 + t^2)^2}
\]

### Step 3: Calculate \( \frac{dy}{dx} \)

Now we have:
\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{2(1 - t^2)}{(1 + t^2)^2}}{\frac{-4t}{(1 + t^2)^2}}
\]

The \( (1 + t^2)^2 \) terms cancel:
\[
\frac{dy}{dx} = \frac{2(1 - t^2)}{-4t} = -\frac{1 - t^2}{2t}
\]

Thus,
\[
\frac{dy}{dx} = -\frac{1 - t^2}{2t}
\]

Would you like additional details on the steps, or do you have any questions?

---

### Additional Questions

1. What happens to \( \frac{dy}{dx} \) as \( t \to 0 \)?
2. How can we confirm that \( x \) and \( y \) describe a conic section?
3. How would \( \frac{dy}{dx} \) change if the parametric equations were different?
4. What are some real-world applications of parametric differentiation?
5. How do you find higher-order derivatives \( \frac{d^2y}{dx^2} \) for parametric equations?

### Tip
When differentiating parametric equations, always ensure the chain rule and quotient rule are applied carefully to avoid mistakes.

If x = (1 - t^2)/(1 + t^2) and y = 2t/(1 + t^2) then find dy/dx

Explore how to find dy/dx for the parametric equations x = (1 - t^2)/(1 + t^2) and y = 2t/(1 + t^2) by using differentiation techniques. This problem demonstrates key concepts in parametric differentiation, including the chain rule and quotient rule, suitable for advanced high school students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation