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use implicit differentiation to find the slope of the tangent line to the curve of x^2/3+y^2/3=4 at the point (-1,3sqrt3)Asked by Nancy
use implicit differentiation to find the slope of the tangent line to the curve of x^2/3+y^2/3=4 at the point (-1,3sqrt3)
Answers
Answered by
bobpursley
taking the first
2/3 x dx + 2/3 ydy=0
dy/dx=-x/y at the point, 1/(3sqrt3)
2/3 x dx + 2/3 ydy=0
dy/dx=-x/y at the point, 1/(3sqrt3)
Answered by
Brittany
((x^2)/3)+((y^2)/3)=4
d/dx 2x/3 + d/dx 2y/3 (y')= 0
y'= -(2x/3)-(2y/3)
at the point (-1,3sqrt3)
y'=(-2(-1)/3)-(2(3sqrt3)/3)=(2/3)-((6sqrt3)/3)=(2/3)-2sqrt3
Answer is (2/3)-2sqrt3 correct?
d/dx 2x/3 + d/dx 2y/3 (y')= 0
y'= -(2x/3)-(2y/3)
at the point (-1,3sqrt3)
y'=(-2(-1)/3)-(2(3sqrt3)/3)=(2/3)-((6sqrt3)/3)=(2/3)-2sqrt3
Answer is (2/3)-2sqrt3 correct?
Answered by
GanonTEK
Is is x^(2/3) or is it (x^2)/3 ?
If it was the latter you could multiply both sides by 3 to get rid of any fractions straight away.
Brittany, y' = -(2x/3) / (2y/3) = -x/y
not y' = -(2x/3)-(2y/3)
If it was the latter you could multiply both sides by 3 to get rid of any fractions straight away.
Brittany, y' = -(2x/3) / (2y/3) = -x/y
not y' = -(2x/3)-(2y/3)
Answered by
Brittany
it is x^(2/3)
Answered by
Nancy
The problem is (x^(2/3))+(y^(2/3))=4
I'm confused on the answer
I'm confused on the answer
Answered by
Nancy
I got the same answer as Brittany, I just want to make sure if it is correct or not
Answered by
GanonTEK
ok let's see...
x^(2/3) + y^(2/3) = 4
differentiating we get
(2/3)*(x^(-1/3)) + (2/3)*(y^(-1/3)).dy/dx = 0
divide across by (2/3) to get
x^(-1/3) + y^(-1/3).dy/dx = 0
Then replace the negative power with 1/ instead. e.g. x^-1 = 1/x
1/x^(1/3) + (1/y^(1/3)).dy/dx = 0
(1/y^(1/3)).dy/dx = -1/x^(1/3)
dy/dx = -y^(1/3)/x^(1/3) = -(y/x)^1/3
pt (1, 3sqrt3)
dy/dx = -(3sqrt3)^1/3
Bit nasty to be honest.
x^(2/3) + y^(2/3) = 4
differentiating we get
(2/3)*(x^(-1/3)) + (2/3)*(y^(-1/3)).dy/dx = 0
divide across by (2/3) to get
x^(-1/3) + y^(-1/3).dy/dx = 0
Then replace the negative power with 1/ instead. e.g. x^-1 = 1/x
1/x^(1/3) + (1/y^(1/3)).dy/dx = 0
(1/y^(1/3)).dy/dx = -1/x^(1/3)
dy/dx = -y^(1/3)/x^(1/3) = -(y/x)^1/3
pt (1, 3sqrt3)
dy/dx = -(3sqrt3)^1/3
Bit nasty to be honest.
Answered by
Steve
of course, you can simplify that last result to just √3, since it is (3^(3/2))^(1/3) = 3^(1/2)
Also, you dropped/gained a minus sign in there somewhere. The point is in QII, where the slope is positive.
Also, you dropped/gained a minus sign in there somewhere. The point is in QII, where the slope is positive.
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