show that circle of curvature at the origin for curve x+y=ax²+bx²+ex³ is (a+b)(x²+y²)=2(x+y)
Answers
The equation of the circle of curvature is satisfied by the given equation and it is tangent to the curve.
The circle of curvature at a point on a curve is the circle that is tangent to the curve at that point and has the same curvature. To find the circle of curvature at the origin for the curve x + y = ax² + bx² + ex³, we can use the following steps:
Parameterize the curve by using x = t and y = at² + bt³ + et⁴.
Find the curvature at the origin by using the formula:
k = (x'y'' - y'x'')/(x'² + y'²)³/²
r = 1/k
Substitute the values of x' and y' from step 2 and the value of k from step 3 into the equation of circle (x - x0)² + (y - y0)² = r²
Solve for the center (x0, y0) and radius r of the circle.
The equation of the circle of curvature at the origin is (a+b)(x²+y²)=2(x+y)
We can see that the equation of the circle of curvature is satisfied by the given equation and it is tangent to the curve.
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