Asked by jack
find equation of tangent line for x^(1/3) + y^(1/2)=2 at point (1,1)
Answers
Answered by
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
Answered by
Damon
Can other teachers post all right ?
Answered by
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
Answered by
Ms. Sue
I've sent your posts to Bob and Leo. I'll send this one too.
Answered by
Damon
Thanks - I can not really answer this because it will not let me write out derivative method.
Answered by
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.
=(
Answered by
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
Answered by
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
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