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
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
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.

=(
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
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
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