find inflection point

Question

a curve is defined by the parametric equation x = 3sin(t) + 4 and y=3cos(t)-1 for 0 <= 1 <= pi, find any point of inflection in the curve

Answer 1

Method 1:

x = 3sin(t) + 4 ⇒ x - 4 = 3sin(t)

y = 3cos(t) - 1 ⇒ y + 1 = 3cos(t)

(x - 4)² + (y + 1)² = (3sin(t))² + (3cos(t))² = 9sin²(t) + 9cos²(t) = 9(sin²(t) + cos²(t)) = 9 = 3²

(x - 4)² + (y + 1)² = 3² is the equation of a circle with center (4, -1) and radius 3.

However, we additionally have that 0 ≤ t ≤ π.

This is what tells us that we are NOT looking at an entire circle but only half of the circle

(for the entire circle, the interval that t is in would need to have at least a 2π width)

The question then becomes, which half?

0 ≤ t ≤ π ⇒ 0 ≤ sin(t) ≤ 1 ⇒ 0 ≤ 3sin(t) ≤ 3 ⇒ 4 ≤ 3sin(t) + 4 ≤ 7 ⇒ 4 ≤ x ≤ 7

0 ≤ t ≤ π ⇒ -1 ≤ cos(t) ≤ 1 ⇒ -3 ≤ 3cos(t) ≤ 3 ⇒ -4 ≤ 3cos(t) - 1 ≤ 2 ⇒ -4 ≤ y ≤ 2

Notice that the width of the interval of x is only 3 whereas the width of the interval of y is 6.

This tells us that we are only looking at half of the circle in the x-direction.

Specifically, we are looking at the right semi-circle with center (4, -1) and radius 3.



This Desmos graph can confirm the analysis so far.

https://www.desmos.com/calculator/vchgpkxo6x

A circle does not have an inflection point.

I am unfortunately not able to find any other source than Wikipedia at this point :-(

https://en.wikipedia.org/wiki/Tangent#:~:text=Circles%2C%20parabolas%2C%20hyperbolas%20and%20ellipses,each%20period%20of%20the%20sine.

Method 2:

Alternatively, there is a general method to find points of inflection in a curve given by parametric equations. There is a nice explanation provided in stack exchange

https://math.stackexchange.com/questions/487170/find-inflection-points-of-parametric-equations

If you evaluate x''(t)y'(t) - y''(t)x'(t) as described in the link, you will find that it is a nonzero constant for any value of t and that the curve does not have a point of inflection.

Answer 2

The curve is a circle with center (4,-1) and radius 6.

The points of inflection are at (1,-1) and (7,-1)

This answer is clear graphically...at these 2 points the concavity changes.

What I am sure of is that the unit circle has inflection points at the ends of the horizontal diameter and this circle must also.

1st of all, (1,-1) is not in the range; therefore, it cannot be a point of inflection. Sorry about that!

However, a point of inflection is NOT defined by the 2nd derivative; it is also not defined only for functions. A point of inflection is defined as a point at which the curve in question changes concavity (from concave up to concave down or vice-versa). The endpoints of the horizontal diameter of a circle meet that definition. Therefore, (7,-1) IS an inflection point. And it is fairly obvious how to demonstrate that simply by drawing secant lines from the end point of the diameter; this can also be done algebraically.