Asked by lily
find the eqt. of the tangent to the curves at (1,2).
y^3+y^3=9(xy-1)
[ans: y=5x-5]
could some1 show me the calculation work bcus my ans is y=(6x+4)/5.
y^3+y^3=9(xy-1)
[ans: y=5x-5]
could some1 show me the calculation work bcus my ans is y=(6x+4)/5.
Answers
Answered by
Reiny
I will guess that you have a typo and one of those first terms is x^3
x^3 + y^3 = 9xy - 9
3x^2 + 3y^2 dy/dx = 9x dy/dx + 9y
dy/dx(3y^2 - 9x) = 9y - 3x^2
at (1,2)
dy/dx(12-9) = 18 - 3
dy/dx = 15/3 = 5
so equation is y = 5x + b
plug in (1,2)
2 = 5(1) + b
b = -3
tangent equation : y = 5x - 3
Their answer is wrong, since (1,2) doesn't even satisfy their answer equation.
x^3 + y^3 = 9xy - 9
3x^2 + 3y^2 dy/dx = 9x dy/dx + 9y
dy/dx(3y^2 - 9x) = 9y - 3x^2
at (1,2)
dy/dx(12-9) = 18 - 3
dy/dx = 15/3 = 5
so equation is y = 5x + b
plug in (1,2)
2 = 5(1) + b
b = -3
tangent equation : y = 5x - 3
Their answer is wrong, since (1,2) doesn't even satisfy their answer equation.
Answered by
lily
the 1st term is y^3
Answered by
Reiny
Then your equation would simplify to
2y^3 = 9xy - 9
6y^2 dy/dx = 9x dy/dx + 9y
dy/dx(6y^2 - 9x) = 9y
for the given point (1,2)
dy/dx(24-9) = 18
dy/dx = 18/15 = 6/5
Now the slope of the line will be 6/5 , but in your answer it is 5, as I had before.
Now it makes even less sense!
Please check your typing, it is frustrating to keep guessing what the student meant in solving these questions.
2y^3 = 9xy - 9
6y^2 dy/dx = 9x dy/dx + 9y
dy/dx(6y^2 - 9x) = 9y
for the given point (1,2)
dy/dx(24-9) = 18
dy/dx = 18/15 = 6/5
Now the slope of the line will be 6/5 , but in your answer it is 5, as I had before.
Now it makes even less sense!
Please check your typing, it is frustrating to keep guessing what the student meant in solving these questions.
Answered by
lily
you get the same ans like me so the ans is wrong.
Is that right?
Is that right?
Answered by
Reiny
It appears that way, but I still think no textbook would have an equation looking like
y^3 + y^3 .....
y^3 + y^3 .....
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