Asked by Marie
Consider the closed curve in the day plane:
2x^2-2xy+y^3=14
a) show that dy/dx=2y-4x/3y^2-2x (I got this part)
b) find equation lines to the curve when y=2
c) if the point (2.5, k) is on the curve, use part b to find the best approximation of the value of k
2x^2-2xy+y^3=14
a) show that dy/dx=2y-4x/3y^2-2x (I got this part)
b) find equation lines to the curve when y=2
c) if the point (2.5, k) is on the curve, use part b to find the best approximation of the value of k
Answers
Answered by
Marie
Consider the closed curve in the xy plane:
2x^2-2xy+y^3=14
a) show that dy/dx=2y-4x/3y^2-2x (I got this part)
b) find equation lines to the curve when y=2
c) if the point (2.5, k) is on the curve, use part b to find the best approximation of the value of k
2x^2-2xy+y^3=14
a) show that dy/dx=2y-4x/3y^2-2x (I got this part)
b) find equation lines to the curve when y=2
c) if the point (2.5, k) is on the curve, use part b to find the best approximation of the value of k
Answered by
Reiny
Your dy/dx should be
(2y-4x)/(3y^2 - 2x)
(the way you have it, only the -4x is divided by 3y^2 )
so when y = 2 , in the original,
2x^2 - 4x + 8 = 14
2x^2 - 4x - 6 = 0
x^2 - 2x - 3 = 0
(x-3)(x+1) = 0
x = 3 or x = -1
b) you probably meant to say:
Find the equation of the tangent lines when y = 2
so the points of contact of the two tangents are
(3,2) and (-1,2)
at (3,2), slope = dy/dx = (4 - 8)/(12 - 6) = -4/6 = -2/3
equation of tangent:
y-2 = (-2/3)(x-3)
3y - 6 = -2x + 6
2x + 3y = 12 or y = (-2/3)x + 4
follow the same steps for the tangent to the other point
c) Not really sure what they want for that part.
Have you studied differentials?
If the point (2.5 , k) is supposed to be "close" to (3,2), that's quite a spread even if we use differentials
Just subbing in x = 2.5 yields the cubic
y^3 - 5y = -1.5
An appr just taking a few values of y gave me
y = appr 2.4 (y = 2.3732...)
(2y-4x)/(3y^2 - 2x)
(the way you have it, only the -4x is divided by 3y^2 )
so when y = 2 , in the original,
2x^2 - 4x + 8 = 14
2x^2 - 4x - 6 = 0
x^2 - 2x - 3 = 0
(x-3)(x+1) = 0
x = 3 or x = -1
b) you probably meant to say:
Find the equation of the tangent lines when y = 2
so the points of contact of the two tangents are
(3,2) and (-1,2)
at (3,2), slope = dy/dx = (4 - 8)/(12 - 6) = -4/6 = -2/3
equation of tangent:
y-2 = (-2/3)(x-3)
3y - 6 = -2x + 6
2x + 3y = 12 or y = (-2/3)x + 4
follow the same steps for the tangent to the other point
c) Not really sure what they want for that part.
Have you studied differentials?
If the point (2.5 , k) is supposed to be "close" to (3,2), that's quite a spread even if we use differentials
Just subbing in x = 2.5 yields the cubic
y^3 - 5y = -1.5
An appr just taking a few values of y gave me
y = appr 2.4 (y = 2.3732...)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.