Maths questions

Question

Given the curve (x^2)y+ax+by=0 have three inflection points, where a and b are constants.

a) Using implicit differentiation, find y' and y'' in terms of a and b.

b) Find the values of a and b if one of the inflection points is (2, 5/2).



c) With the answer in (b), find another two inflection points.

Answer 1

Even though the problem asks for implicit differentiation, it is easier to rearrange the problem in terms of y and then use the quotient rule of differentiation.

x²y + a x + b y = 0 ------------> y = -ax / (x² + b)

Taking the implicit differentiation, we get:

2 xy + x² y' + a + b y' = 0 --------> (2 xy + a) + (x² + b) y' = 0

y' = - (2 xy + a) / (x² + b) , you can substitute for y = -ax / (x² + b), and then simplify (or just use quotient rule on y = -ax / (x² + b)). This will give same result. Using quotient, we get:

y' = (-a (x² +b) + 2 a x²) / (x² + b)² = a (- x² - b + 2 x²) /(x² + b)² = a (x² - b) / (x² + b)²

y'' = (2ax (x² + b)² - a (x² - b) 4 x (x² + b)) / (x² + b)⁴ , simplifying:

y'' = (6 a b x - 2 a x³) / (x² + b)³

We are given (2, 5/2) as an inflection point, y'' = 0 at this point. Substitute for x = 2 in y'':

0 = (12 ab - 16 a) divide by a

12 b - 16 = 0 ----------------> b = 16/12 = 4/3

We can use the same inflection point and b value in y = -ax / (x² + b):

a = - y (x² + b) / x = - (5/2) (4 + 4/3) / 2 = - 20 / 3

a = - 20 / 3

Now, since we have a and b, we can make y'' = 0 and solve for x:

(6 a b x - 2 a x³) / (x² + b)³ = 0 , multiplying by (x² + b)³ and dividing by a, we get:

6 b x - 2 x³ = 0

6 (4/3) x - 2 x³ = 0

8 x - 2 x³ = 0 divide by 2 and rearrange:

x³ - 4 x= 0

x (x² - 4) = 0

x = 0 and (x² - 4) = 0 -------------------> x = -2 and 2 (we already know (2 , 5/2) inflection point)

x = 0 ---------------> y = -ax / (x² + b) = 0

x = -2 ---------------> y = -ax / (x² + b) = 20 x / (3 x² + 4) = - 40 / 16 = - 5 / 2

Therefore, the other two inflection points are:

(0, 0) and (- 2, - 5 / 2)