Question

Find the radius of curvature at the origin for
2x4 + 3y4 + 4x²y + xy - y2 + 2x = 0 .
Trace the curve y = x3 - (x + 1)2. Also find the asymptotes of the following curve
(x2 - y2)2 – 4x² + x = 0.

Answer 1

The radius of curvature at the origin for the equation 2x^4 + 3y^4 + 4x²y + xy - y^2 + 2x = 0 is 1/20*((13)^(3/2)).To trace the curve y = x^3 - (x + 1)^2, we need to plot the points (x, y) that satisfy the equation.The asymptotes of the curve (x^2 - y^2)^2 – 4x² + x = 0 are the x-axis and the line x = 4.

To Find:

• radius of curvature at the origin for the first equation
• points that satisfy the equation y = x^3 - (x + 1)^2
• asymptotes of the curve (x^2 - y^2)^2 – 4x² + x = 0.

Given:

• equation 2x^4 + 3y^4 + 4x²y + xy - y^2 + 2x = 0
• equation y = x^3 - (x + 1)^2
• equation (x^2 - y^2)^2 – 4x² + x = 0

Solution:

The radius of curvature at the origin for 2x^4 + 3y^4 + 4x²y + xy - y^2 + 2x = 0 can be found by using the formula

1/R = (4Fx + 2Gx + 2Fy + 4Gy)/((1+2Fx^2 + 2Gx^2)^(3/2))

Where F = 2x^4 + 4xy + 2x, G = 3y^4 - y^2 + 2, x = y = 0

By substituting values we get 1/R = (42 + 22 + 23 + 42)/((1+22^2 + 23^2)^(3/2)) = 20/((1+12)^(3/2)) = 20/((13)^(3/2))

The radius of curvature at the origin is 1/20*((13)^(3/2))

To trace the curve y = x^3 - (x + 1)^2, we need to plot the points (x, y) that satisfy the equation.

For the curve (x^2 - y^2)^2 – 4x² + x = 0.

The asymptotes are the lines that the graph of the function approaches but never touches.

The equation of the asymptotes can be found by setting the denominator equal to zero and solving for x and y.

(x^2 - y^2)^2 = 4x² - x = 0

solving for x we get x(x-4) = 0, x = 0 and x = 4.

solving for y we get y = ± √x^2.

The asymptotes of the curve (x^2 - y^2)^2 – 4x² + x = 0 are the x-axis and the line x = 4.

Therefore,The radius of curvature at the origin for the equation 2x^4 + 3y^4 + 4x²y + xy - y^2 + 2x = 0 is 1/20*((13)^(3/2)).To trace the curve y = x^3 - (x + 1)^2, we need to plot the points (x, y) that satisfy the equation.The asymptotes of the curve (x^2 - y^2)^2 – 4x² + x = 0 are the x-axis and the line x = 4.

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Answer 2

The curve has two pairs of asymptotes, namely y =$(2x - 1)√x, y$ = -$(2x - 1)√x, x = 0$, and x = 1/4.

To trace the curve y = x³ - (x + 1)², we can eliminate y from the given equation 2x4 + 3y4 + 4x²y + xy - y2 + 2x = 0 by substituting y with x³ - (x + 1)².

So, we get:

2x⁴ + 3(x³ - (x + 1)²)⁴ + 4x²(x³ - (x + 1)²) + x(x³ - (x + 1)²) - (x³ - (x + 1)²)² + 2x = 0

Simplifying the above equation, we get:

10x⁷ + 2x⁶ - 17x⁵ - 36x⁴ + 8x³ + 33x² - 4x - 1 = 0

Using numerical methods or a graphing calculator, we can plot the curve given by this equation, which is the same as the curve y = x³ - (x + 1)².

To find the asymptotes of the curve (x² - y²)² - 4x² + x = 0, we can rewrite the equation as:

(x² - y²)² = 4x² - x

Taking the square root on both sides, we get:

x² - y² =$±√(4x² - x)$

We can simplify the above equation using the identity a² - b² = (a + b)(a - b), as follows:

(x + y)(x - y) =$±√x(4x - 1)$

Therefore, the asymptotes of the curve are given by the equations:

(x + y)(x - y) = $√x(4x - 1) and (x + y)(x - y) = -√x(4x - 1)$

Simplifying further, we get:

y = $±(2x - 1)√x$

and

y = $∓(2x - 1)√x$

So, the curve has two pairs of asymptotes, namely y =$(2x - 1)√x, y$ = -$(2x - 1)√x, x = 0$, and x = 1/4.

For such more questions on curve,

https://brainly.in/question/37734131

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