Line Integral Work Done [closed]
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I have a problem when comes to question 2, I don't know how to put this into parametric form. And I am not sure if this is a parabola. Thanks in advance.
integration parametric line-integrals
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closed as off-topic by Saad, Abcd, Shailesh, Lee David Chung Lin, user91500 Jan 12 at 9:58
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$begingroup$
I have a problem when comes to question 2, I don't know how to put this into parametric form. And I am not sure if this is a parabola. Thanks in advance.
integration parametric line-integrals
$endgroup$
closed as off-topic by Saad, Abcd, Shailesh, Lee David Chung Lin, user91500 Jan 12 at 9:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Abcd, Shailesh, Lee David Chung Lin, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I have a problem when comes to question 2, I don't know how to put this into parametric form. And I am not sure if this is a parabola. Thanks in advance.
integration parametric line-integrals
$endgroup$
I have a problem when comes to question 2, I don't know how to put this into parametric form. And I am not sure if this is a parabola. Thanks in advance.
integration parametric line-integrals
integration parametric line-integrals
asked Jan 11 at 22:51
joshjosh
155
155
closed as off-topic by Saad, Abcd, Shailesh, Lee David Chung Lin, user91500 Jan 12 at 9:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Abcd, Shailesh, Lee David Chung Lin, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Abcd, Shailesh, Lee David Chung Lin, user91500 Jan 12 at 9:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Abcd, Shailesh, Lee David Chung Lin, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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If $y=4x^2$ then the $y$-coordinate of 4 times the square of the $x$-coordinate.
Let the $x$-coordinate be $t$, then the $y$-coordinate of $4t^2$.
Your path is given by $gamma(t) = (t,4t^2)$, where $0 le t le 1$.
The curve $y=4x^2$ is indeed a parabola, with focus $(x,y)=(0,frac{1}{16})$ and directrix $y=-frac{1}{16}$.
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $y=4x^2$ then the $y$-coordinate of 4 times the square of the $x$-coordinate.
Let the $x$-coordinate be $t$, then the $y$-coordinate of $4t^2$.
Your path is given by $gamma(t) = (t,4t^2)$, where $0 le t le 1$.
The curve $y=4x^2$ is indeed a parabola, with focus $(x,y)=(0,frac{1}{16})$ and directrix $y=-frac{1}{16}$.
$endgroup$
add a comment |
$begingroup$
If $y=4x^2$ then the $y$-coordinate of 4 times the square of the $x$-coordinate.
Let the $x$-coordinate be $t$, then the $y$-coordinate of $4t^2$.
Your path is given by $gamma(t) = (t,4t^2)$, where $0 le t le 1$.
The curve $y=4x^2$ is indeed a parabola, with focus $(x,y)=(0,frac{1}{16})$ and directrix $y=-frac{1}{16}$.
$endgroup$
add a comment |
$begingroup$
If $y=4x^2$ then the $y$-coordinate of 4 times the square of the $x$-coordinate.
Let the $x$-coordinate be $t$, then the $y$-coordinate of $4t^2$.
Your path is given by $gamma(t) = (t,4t^2)$, where $0 le t le 1$.
The curve $y=4x^2$ is indeed a parabola, with focus $(x,y)=(0,frac{1}{16})$ and directrix $y=-frac{1}{16}$.
$endgroup$
If $y=4x^2$ then the $y$-coordinate of 4 times the square of the $x$-coordinate.
Let the $x$-coordinate be $t$, then the $y$-coordinate of $4t^2$.
Your path is given by $gamma(t) = (t,4t^2)$, where $0 le t le 1$.
The curve $y=4x^2$ is indeed a parabola, with focus $(x,y)=(0,frac{1}{16})$ and directrix $y=-frac{1}{16}$.
edited Jan 11 at 23:18
answered Jan 11 at 23:11
Fly by NightFly by Night
25.9k32978
25.9k32978
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