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Help please.......................... ​

Help please.......................... ​-example-1
User Maztch
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1 Answer

4 votes

Answer:

-dx/dy = (1 + x) √(1 − x²)

Explanation:

y = √((1 − x) / (1 + x))

Squaring both sides:

y² = (1 − x) / (1 + x)

Take derivative of both sides (use power rule and chain rule on the left, and quotient rule on the right):

2y dy/dx = [(1 + x)(-1) − (1 − x)(1)] / (1 + x)²

2y dy/dx = (-1 − x − 1 + x) / (1 + x)²

2y dy/dx = -2 / (1 + x)²

y dy/dx = -1 / (1 + x)²

Substitute the expression for y:

√((1 − x) / (1 + x)) dy/dx = -1 / (1 + x)²

Multiply both sides by 1 + x:

√((1 − x) (1 + x)) dy/dx = -1 / (1 + x)

√(1 − x²) dy/dx = -1 / (1 + x)

Solve for dy/dx:

dy/dx = -1 / [ (1 + x) √(1 − x²) ]

The gradient of the normal is the slope of the perpendicular line:

-dx/dy = (1 + x) √(1 − x²)

Here's a graph:

desmos.com/calculator/kbglyjdzaj

User Vbranden
by
6.1k points