Question
find equation of tangent line for x^(1/3) + y^(1/2)=2 at point (1,1)
Answers
Damon
Once again I can not post my answer but in the end I get
3 y = -2 x + 5
3 y = -2 x + 5
Damon
Can other teachers post all right ?
Damon
no luck - hope someone can help you.
the idea is to do f(d+dx) - f(x)
where f(x+dx) = f(x) + (dy/dx) dx
the idea is to do f(d+dx) - f(x)
where f(x+dx) = f(x) + (dy/dx) dx
Ms. Sue
I've sent your posts to Bob and Leo. I'll send this one too.
Damon
Thanks - I can not really answer this because it will not let me write out derivative method.
Writeacher
Bob knows. He has cleaned out the banned list, and said he'd check more specifically later.
I've texted Leo ... I'm surprised there's been no answer.
=(
I've texted Leo ... I'm surprised there's been no answer.
=(
Reiny
(1/3)x^(-2/3) + (1/2)y^(-1/2) dy/dx = 0
at (1,1)
(1/3)(1) + (1/2)(1)dy/dx =0
dy/dx =-(1/2) / (1/3) = -3/2
so y-1 = (-3/2)(x-1)
2y - 2 = -3x + 3
2y = -3x + 5
y = (-3/2)x + 5/2
Damon, I had the same problem, now it seems to work
at (1,1)
(1/3)(1) + (1/2)(1)dy/dx =0
dy/dx =-(1/2) / (1/3) = -3/2
so y-1 = (-3/2)(x-1)
2y - 2 = -3x + 3
2y = -3x + 5
y = (-3/2)x + 5/2
Damon, I had the same problem, now it seems to work
Damon
tangent line for x^(1/3) + y^(1/2)=2 at point (1,1)
-----------------------------------
(1/3) / x^(2/3) + (1/2) dy/dx /y^(1/2)=0
so
(1/2)dy/dx = -(1/3) y^(1/2)/x^(2/3)
at (1.1)
dy/dx = -(2/3)
then
y = -2x/3 + b
1 = -2/3 + b
b = 5/3
so
y = -2x/3 + 5/3
3 y = -2 x + 5
-----------------------------------
(1/3) / x^(2/3) + (1/2) dy/dx /y^(1/2)=0
so
(1/2)dy/dx = -(1/3) y^(1/2)/x^(2/3)
at (1.1)
dy/dx = -(2/3)
then
y = -2x/3 + b
1 = -2/3 + b
b = 5/3
so
y = -2x/3 + 5/3
3 y = -2 x + 5