Mixing
Function $ f(x,y)=x+frac{y^3}{3} $ cut the xy-plane in a cutting-curve $h$. Find the tangent fuction of h in...
0
$begingroup$
$$ f(x,y)=x+frac{y^3}{3} , D(f)=text{(x,y)}in R |(x^2 +y^2le2) $$
So what i thinking here is find the function h, then find the tangent. But i dont know how to do or is there another way? Thank you for reading
calculus 3d curves
asked Jan 24 at 3:44
Vu Thanh PhanVu Thanh Phan
347
$endgroup$
add a comment |
0
$begingroup$
$$ f(x,y)=x+frac{y^3}{3} , D(f)=text{(x,y)}in R |(x^2 +y^2le2) $$
So what i thinking here is find the function h, then find the tangent. But i dont know how to do or is there another way? Thank you for reading
calculus 3d curves
asked Jan 24 at 3:44
Vu Thanh PhanVu Thanh Phan
347
$endgroup$
$begingroup$
How does f cut the plane? What is h?
$endgroup$
– William Elliot
Jan 24 at 4:46
$begingroup$
Sorry i typed wrong. Just editted that to f(x,y). h is the fuction of the curve which is formed by the cutting of f(x,y) and the xy-plane
$endgroup$
– Vu Thanh Phan
Jan 24 at 4:53
$begingroup$
So the curve is 3x + y$^3$ = 0? What is D(f) supposed to be?
$endgroup$
– William Elliot
Jan 24 at 8:57
$begingroup$
D(f) mean the definition range of the function f. Because it is only a part of the excercise and the other part need the definition range to be completed.
$endgroup$
– Vu Thanh Phan
Jan 24 at 18:44
$begingroup$
Thanks for your answer. Can you help me more with how to find the slope in order to find tangent function of h:= 3x+y^3=0?
$endgroup$
– Vu Thanh Phan
Jan 24 at 18:45
add a comment |
0
0
0
$begingroup$
$$ f(x,y)=x+frac{y^3}{3} , D(f)=text{(x,y)}in R |(x^2 +y^2le2) $$
So what i thinking here is find the function h, then find the tangent. But i dont know how to do or is there another way? Thank you for reading
calculus 3d curves
asked Jan 24 at 3:44
Vu Thanh PhanVu Thanh Phan
347
$endgroup$
$$ f(x,y)=x+frac{y^3}{3} , D(f)=text{(x,y)}in R |(x^2 +y^2le2) $$
So what i thinking here is find the function h, then find the tangent. But i dont know how to do or is there another way? Thank you for reading
calculus 3d curves
calculus 3d curves
asked Jan 24 at 3:44
Vu Thanh PhanVu Thanh Phan
347
asked Jan 24 at 3:44
Vu Thanh PhanVu Thanh Phan
347
edited Jan 24 at 4:52
Vu Thanh Phan
asked Jan 24 at 3:44
Vu Thanh PhanVu Thanh Phan
347
asked Jan 24 at 3:44
Vu Thanh PhanVu Thanh Phan
347
asked Jan 24 at 3:44
Vu Thanh PhanVu Thanh Phan
347
347
$begingroup$
How does f cut the plane? What is h?
$endgroup$
– William Elliot
Jan 24 at 4:46
$begingroup$
Sorry i typed wrong. Just editted that to f(x,y). h is the fuction of the curve which is formed by the cutting of f(x,y) and the xy-plane
$endgroup$
– Vu Thanh Phan
Jan 24 at 4:53
$begingroup$
So the curve is 3x + y$^3$ = 0? What is D(f) supposed to be?
$endgroup$
– William Elliot
Jan 24 at 8:57
$begingroup$
D(f) mean the definition range of the function f. Because it is only a part of the excercise and the other part need the definition range to be completed.
$endgroup$
– Vu Thanh Phan
Jan 24 at 18:44
$begingroup$
Thanks for your answer. Can you help me more with how to find the slope in order to find tangent function of h:= 3x+y^3=0?
$endgroup$
– Vu Thanh Phan
Jan 24 at 18:45
add a comment |
$begingroup$
How does f cut the plane? What is h?
$endgroup$
– William Elliot
Jan 24 at 4:46
$begingroup$
Sorry i typed wrong. Just editted that to f(x,y). h is the fuction of the curve which is formed by the cutting of f(x,y) and the xy-plane
$endgroup$
– Vu Thanh Phan
Jan 24 at 4:53
$begingroup$
So the curve is 3x + y$^3$ = 0? What is D(f) supposed to be?
$endgroup$
– William Elliot
Jan 24 at 8:57
$begingroup$
D(f) mean the definition range of the function f. Because it is only a part of the excercise and the other part need the definition range to be completed.
$endgroup$
– Vu Thanh Phan
Jan 24 at 18:44
$begingroup$
Thanks for your answer. Can you help me more with how to find the slope in order to find tangent function of h:= 3x+y^3=0?
$endgroup$
– Vu Thanh Phan
Jan 24 at 18:45
$begingroup$
How does f cut the plane? What is h?
$endgroup$
– William Elliot
Jan 24 at 4:46
$begingroup$
How does f cut the plane? What is h?
$endgroup$
– William Elliot
Jan 24 at 4:46
$begingroup$
Sorry i typed wrong. Just editted that to f(x,y). h is the fuction of the curve which is formed by the cutting of f(x,y) and the xy-plane
$endgroup$
– Vu Thanh Phan
Jan 24 at 4:53
$begingroup$
Sorry i typed wrong. Just editted that to f(x,y). h is the fuction of the curve which is formed by the cutting of f(x,y) and the xy-plane
$endgroup$
– Vu Thanh Phan
Jan 24 at 4:53
$begingroup$
So the curve is 3x + y$^3$ = 0? What is D(f) supposed to be?
$endgroup$
– William Elliot
Jan 24 at 8:57
$begingroup$
So the curve is 3x + y$^3$ = 0? What is D(f) supposed to be?
$endgroup$
– William Elliot
Jan 24 at 8:57
$begingroup$
D(f) mean the definition range of the function f. Because it is only a part of the excercise and the other part need the definition range to be completed.
$endgroup$
– Vu Thanh Phan
Jan 24 at 18:44
$begingroup$
D(f) mean the definition range of the function f. Because it is only a part of the excercise and the other part need the definition range to be completed.
$endgroup$
– Vu Thanh Phan
Jan 24 at 18:44
$begingroup$
Thanks for your answer. Can you help me more with how to find the slope in order to find tangent function of h:= 3x+y^3=0?
$endgroup$
– Vu Thanh Phan
Jan 24 at 18:45
$begingroup$
Thanks for your answer. Can you help me more with how to find the slope in order to find tangent function of h:= 3x+y^3=0?
$endgroup$
– Vu Thanh Phan
Jan 24 at 18:45
add a comment |
1 Answer
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$begingroup$
So you want to find the tangent of 3x + y$^3$ = 0 at (9,-3).
As 3 + 3y$^2$y' = 0, y'(x) = -/y$^2$.
At the given point, y'(9) = -1/9 = m.
The tangent is y + 3 = m(x - 9).
answered Jan 25 at 3:55
William ElliotWilliam Elliot
8,6922720
$endgroup$
add a comment |
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1 Answer
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1 Answer
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1
$begingroup$
So you want to find the tangent of 3x + y$^3$ = 0 at (9,-3).
As 3 + 3y$^2$y' = 0, y'(x) = -/y$^2$.
At the given point, y'(9) = -1/9 = m.
The tangent is y + 3 = m(x - 9).
answered Jan 25 at 3:55
William ElliotWilliam Elliot
8,6922720
$endgroup$
add a comment |
1
$begingroup$
So you want to find the tangent of 3x + y$^3$ = 0 at (9,-3).
As 3 + 3y$^2$y' = 0, y'(x) = -/y$^2$.
At the given point, y'(9) = -1/9 = m.
The tangent is y + 3 = m(x - 9).
answered Jan 25 at 3:55
William ElliotWilliam Elliot
8,6922720
$endgroup$
add a comment |
1
1
1
$begingroup$
So you want to find the tangent of 3x + y$^3$ = 0 at (9,-3).
As 3 + 3y$^2$y' = 0, y'(x) = -/y$^2$.
At the given point, y'(9) = -1/9 = m.
The tangent is y + 3 = m(x - 9).
answered Jan 25 at 3:55
William ElliotWilliam Elliot
8,6922720
$endgroup$
So you want to find the tangent of 3x + y$^3$ = 0 at (9,-3).
As 3 + 3y$^2$y' = 0, y'(x) = -/y$^2$.
At the given point, y'(9) = -1/9 = m.
The tangent is y + 3 = m(x - 9).
answered Jan 25 at 3:55
William ElliotWilliam Elliot
8,6922720
answered Jan 25 at 3:55
William ElliotWilliam Elliot
8,6922720
answered Jan 25 at 3:55
William ElliotWilliam Elliot
8,6922720
answered Jan 25 at 3:55
William ElliotWilliam Elliot
8,6922720
8,6922720
add a comment |
add a comment |
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$begingroup$
How does f cut the plane? What is h?
$endgroup$
– William Elliot
Jan 24 at 4:46
$begingroup$
Sorry i typed wrong. Just editted that to f(x,y). h is the fuction of the curve which is formed by the cutting of f(x,y) and the xy-plane
$endgroup$
– Vu Thanh Phan
Jan 24 at 4:53
$begingroup$
So the curve is 3x + y$^3$ = 0? What is D(f) supposed to be?
$endgroup$
– William Elliot
Jan 24 at 8:57
$begingroup$
D(f) mean the definition range of the function f. Because it is only a part of the excercise and the other part need the definition range to be completed.
$endgroup$
– Vu Thanh Phan
Jan 24 at 18:44
$begingroup$
Thanks for your answer. Can you help me more with how to find the slope in order to find tangent function of h:= 3x+y^3=0?
$endgroup$
– Vu Thanh Phan
Jan 24 at 18:45