Line integral with vector field in polar coordinates
$begingroup$
I have the following problem:
Given a vector field in polar coordinates $$ mathbf{F}(r,theta) = -4 sin theta mathbf{i} + 4 sin theta mathbf{j},$$ calculate the work done when a particle is moved from point $(1,0)$ to the origin, following the spiral whose polar equation is $r = e^{-theta}.$
My attempt was to write the equation of the spiral like so $$mathbf{alpha}(t) = e^{-theta} cos theta mathbf{i} + e^{-theta} sin theta mathbf{j} \ mathbf{alpha}'(t) = -e^{-theta}(cos theta + sin theta mathbf{i} + sin theta - cos theta mathbf{j}),$$ so the line integral would become $$int_C mathbf{F}cdotmathbf{alpha'}(t) = int_C 8 e^{-theta}sinthetacostheta dtheta.$$
But this doesn't give me the right answer, what am I doing wrong?
NOTE: I know this question was asked before, but it doesn't have an accepted answer, and what I read from there wasn't very helpful.
multivariable-calculus
edited Dec 26 '18 at 19:08
mechanodroid
28.7k62548
asked Mar 10 '17 at 1:19
lorenzattractorlorenzattractor
9212
$endgroup$
add a comment |
$begingroup$
I have the following problem:
Given a vector field in polar coordinates $$ mathbf{F}(r,theta) = -4 sin theta mathbf{i} + 4 sin theta mathbf{j},$$ calculate the work done when a particle is moved from point $(1,0)$ to the origin, following the spiral whose polar equation is $r = e^{-theta}.$
My attempt was to write the equation of the spiral like so $$mathbf{alpha}(t) = e^{-theta} cos theta mathbf{i} + e^{-theta} sin theta mathbf{j} \ mathbf{alpha}'(t) = -e^{-theta}(cos theta + sin theta mathbf{i} + sin theta - cos theta mathbf{j}),$$ so the line integral would become $$int_C mathbf{F}cdotmathbf{alpha'}(t) = int_C 8 e^{-theta}sinthetacostheta dtheta.$$
But this doesn't give me the right answer, what am I doing wrong?
NOTE: I know this question was asked before, but it doesn't have an accepted answer, and what I read from there wasn't very helpful.
multivariable-calculus
edited Dec 26 '18 at 19:08
mechanodroid
28.7k62548
asked Mar 10 '17 at 1:19
lorenzattractorlorenzattractor
9212
$endgroup$
2
$begingroup$
Did you integrate from $0$ to $infty$? What is the correct answer according to the book?
$endgroup$
– Ofek Gillon
Mar 10 '17 at 15:33
$begingroup$
No, I haven't tried integrating from $0$ to $infty$, I'll try that right away. The book says that the answer is $frac{8}{5}.$
$endgroup$
– lorenzattractor
Mar 11 '17 at 0:08
$begingroup$
It worked! I integrated from $0$ to $infty$ and got the correct answer, but why? I really don't get it why it worked.
$endgroup$
– lorenzattractor
Mar 11 '17 at 0:48
$begingroup$
Because you need to integrate from $(1,0)$ to $(0,0)$. We see that for $theta = 0 $ we get the point $(1,0)$, but in which $theta$ do we approach $(0,0)$? You can see that you need to solve the equation $e^{-theta} = 0 Rightarrow theta= infty$ , so you need to integrate from $theta=0$ to $theta=infty$
$endgroup$
– Ofek Gillon
Mar 11 '17 at 7:50
add a comment |
$begingroup$
I have the following problem:
Given a vector field in polar coordinates $$ mathbf{F}(r,theta) = -4 sin theta mathbf{i} + 4 sin theta mathbf{j},$$ calculate the work done when a particle is moved from point $(1,0)$ to the origin, following the spiral whose polar equation is $r = e^{-theta}.$
My attempt was to write the equation of the spiral like so $$mathbf{alpha}(t) = e^{-theta} cos theta mathbf{i} + e^{-theta} sin theta mathbf{j} \ mathbf{alpha}'(t) = -e^{-theta}(cos theta + sin theta mathbf{i} + sin theta - cos theta mathbf{j}),$$ so the line integral would become $$int_C mathbf{F}cdotmathbf{alpha'}(t) = int_C 8 e^{-theta}sinthetacostheta dtheta.$$
But this doesn't give me the right answer, what am I doing wrong?
NOTE: I know this question was asked before, but it doesn't have an accepted answer, and what I read from there wasn't very helpful.
multivariable-calculus
edited Dec 26 '18 at 19:08
mechanodroid
28.7k62548
asked Mar 10 '17 at 1:19
lorenzattractorlorenzattractor
9212
$endgroup$
I have the following problem:
Given a vector field in polar coordinates $$ mathbf{F}(r,theta) = -4 sin theta mathbf{i} + 4 sin theta mathbf{j},$$ calculate the work done when a particle is moved from point $(1,0)$ to the origin, following the spiral whose polar equation is $r = e^{-theta}.$
My attempt was to write the equation of the spiral like so $$mathbf{alpha}(t) = e^{-theta} cos theta mathbf{i} + e^{-theta} sin theta mathbf{j} \ mathbf{alpha}'(t) = -e^{-theta}(cos theta + sin theta mathbf{i} + sin theta - cos theta mathbf{j}),$$ so the line integral would become $$int_C mathbf{F}cdotmathbf{alpha'}(t) = int_C 8 e^{-theta}sinthetacostheta dtheta.$$
But this doesn't give me the right answer, what am I doing wrong?
NOTE: I know this question was asked before, but it doesn't have an accepted answer, and what I read from there wasn't very helpful.
multivariable-calculus
multivariable-calculus
edited Dec 26 '18 at 19:08
mechanodroid
28.7k62548
asked Mar 10 '17 at 1:19
lorenzattractorlorenzattractor
9212
edited Dec 26 '18 at 19:08
mechanodroid
28.7k62548
asked Mar 10 '17 at 1:19
lorenzattractorlorenzattractor
9212
edited Dec 26 '18 at 19:08
mechanodroid
28.7k62548
edited Dec 26 '18 at 19:08
mechanodroid
28.7k62548
edited Dec 26 '18 at 19:08
mechanodroid
28.7k62548
28.7k62548
asked Mar 10 '17 at 1:19
lorenzattractorlorenzattractor
9212
asked Mar 10 '17 at 1:19
lorenzattractorlorenzattractor
9212
asked Mar 10 '17 at 1:19
lorenzattractorlorenzattractor
9212
9212
2
$begingroup$
Did you integrate from $0$ to $infty$? What is the correct answer according to the book?
$endgroup$
– Ofek Gillon
Mar 10 '17 at 15:33
$begingroup$
No, I haven't tried integrating from $0$ to $infty$, I'll try that right away. The book says that the answer is $frac{8}{5}.$
$endgroup$
– lorenzattractor
Mar 11 '17 at 0:08
$begingroup$
It worked! I integrated from $0$ to $infty$ and got the correct answer, but why? I really don't get it why it worked.
$endgroup$
– lorenzattractor
Mar 11 '17 at 0:48
$begingroup$
Because you need to integrate from $(1,0)$ to $(0,0)$. We see that for $theta = 0 $ we get the point $(1,0)$, but in which $theta$ do we approach $(0,0)$? You can see that you need to solve the equation $e^{-theta} = 0 Rightarrow theta= infty$ , so you need to integrate from $theta=0$ to $theta=infty$
$endgroup$
– Ofek Gillon
Mar 11 '17 at 7:50
add a comment |
2
$begingroup$
Did you integrate from $0$ to $infty$? What is the correct answer according to the book?
$endgroup$
– Ofek Gillon
Mar 10 '17 at 15:33
$begingroup$
No, I haven't tried integrating from $0$ to $infty$, I'll try that right away. The book says that the answer is $frac{8}{5}.$
$endgroup$
– lorenzattractor
Mar 11 '17 at 0:08
$begingroup$
It worked! I integrated from $0$ to $infty$ and got the correct answer, but why? I really don't get it why it worked.
$endgroup$
– lorenzattractor
Mar 11 '17 at 0:48
$begingroup$
Because you need to integrate from $(1,0)$ to $(0,0)$. We see that for $theta = 0 $ we get the point $(1,0)$, but in which $theta$ do we approach $(0,0)$? You can see that you need to solve the equation $e^{-theta} = 0 Rightarrow theta= infty$ , so you need to integrate from $theta=0$ to $theta=infty$
$endgroup$
– Ofek Gillon
Mar 11 '17 at 7:50
2
2
$begingroup$
Did you integrate from $0$ to $infty$? What is the correct answer according to the book?
$endgroup$
– Ofek Gillon
Mar 10 '17 at 15:33
$begingroup$
Did you integrate from $0$ to $infty$? What is the correct answer according to the book?
$endgroup$
– Ofek Gillon
Mar 10 '17 at 15:33
$begingroup$
No, I haven't tried integrating from $0$ to $infty$, I'll try that right away. The book says that the answer is $frac{8}{5}.$
$endgroup$
– lorenzattractor
Mar 11 '17 at 0:08
$begingroup$
No, I haven't tried integrating from $0$ to $infty$, I'll try that right away. The book says that the answer is $frac{8}{5}.$
$endgroup$
– lorenzattractor
Mar 11 '17 at 0:08
$begingroup$
It worked! I integrated from $0$ to $infty$ and got the correct answer, but why? I really don't get it why it worked.
$endgroup$
– lorenzattractor
Mar 11 '17 at 0:48
$begingroup$
It worked! I integrated from $0$ to $infty$ and got the correct answer, but why? I really don't get it why it worked.
$endgroup$
– lorenzattractor
Mar 11 '17 at 0:48
$begingroup$
Because you need to integrate from $(1,0)$ to $(0,0)$. We see that for $theta = 0 $ we get the point $(1,0)$, but in which $theta$ do we approach $(0,0)$? You can see that you need to solve the equation $e^{-theta} = 0 Rightarrow theta= infty$ , so you need to integrate from $theta=0$ to $theta=infty$
$endgroup$
– Ofek Gillon
Mar 11 '17 at 7:50
$begingroup$
Because you need to integrate from $(1,0)$ to $(0,0)$. We see that for $theta = 0 $ we get the point $(1,0)$, but in which $theta$ do we approach $(0,0)$? You can see that you need to solve the equation $e^{-theta} = 0 Rightarrow theta= infty$ , so you need to integrate from $theta=0$ to $theta=infty$
$endgroup$
– Ofek Gillon
Mar 11 '17 at 7:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As Ofek pointed out, one should integrate from $0$ to $infty$ given that solving the equation $e^{- theta} = 0$ gives $ theta = infty$, and this is true because we are using the polar equation given in the exercise.
So the answer to this problem is simply taking $mathbf{F}(r,theta)$, compute the dot product with the derivative of the parametric curve $alpha ' (t),$ and finally compute the following integral using integration by parts $$int_0^{infty} 4 e^{-theta} sin 2theta d theta,$$ where I have used the trigonometric identity $2 sin theta cos theta = sin (2 theta).$ This will yield the correct answer, and I credit Ofek for it.
answered Aug 15 '17 at 2:39
lorenzattractorlorenzattractor
9212
$endgroup$
add a comment |
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$begingroup$
As Ofek pointed out, one should integrate from $0$ to $infty$ given that solving the equation $e^{- theta} = 0$ gives $ theta = infty$, and this is true because we are using the polar equation given in the exercise.
So the answer to this problem is simply taking $mathbf{F}(r,theta)$, compute the dot product with the derivative of the parametric curve $alpha ' (t),$ and finally compute the following integral using integration by parts $$int_0^{infty} 4 e^{-theta} sin 2theta d theta,$$ where I have used the trigonometric identity $2 sin theta cos theta = sin (2 theta).$ This will yield the correct answer, and I credit Ofek for it.
answered Aug 15 '17 at 2:39
lorenzattractorlorenzattractor
9212
$endgroup$
add a comment |
$begingroup$
As Ofek pointed out, one should integrate from $0$ to $infty$ given that solving the equation $e^{- theta} = 0$ gives $ theta = infty$, and this is true because we are using the polar equation given in the exercise.
So the answer to this problem is simply taking $mathbf{F}(r,theta)$, compute the dot product with the derivative of the parametric curve $alpha ' (t),$ and finally compute the following integral using integration by parts $$int_0^{infty} 4 e^{-theta} sin 2theta d theta,$$ where I have used the trigonometric identity $2 sin theta cos theta = sin (2 theta).$ This will yield the correct answer, and I credit Ofek for it.
answered Aug 15 '17 at 2:39
lorenzattractorlorenzattractor
9212
$endgroup$
add a comment |
$begingroup$
As Ofek pointed out, one should integrate from $0$ to $infty$ given that solving the equation $e^{- theta} = 0$ gives $ theta = infty$, and this is true because we are using the polar equation given in the exercise.
So the answer to this problem is simply taking $mathbf{F}(r,theta)$, compute the dot product with the derivative of the parametric curve $alpha ' (t),$ and finally compute the following integral using integration by parts $$int_0^{infty} 4 e^{-theta} sin 2theta d theta,$$ where I have used the trigonometric identity $2 sin theta cos theta = sin (2 theta).$ This will yield the correct answer, and I credit Ofek for it.
answered Aug 15 '17 at 2:39
lorenzattractorlorenzattractor
9212
$endgroup$
As Ofek pointed out, one should integrate from $0$ to $infty$ given that solving the equation $e^{- theta} = 0$ gives $ theta = infty$, and this is true because we are using the polar equation given in the exercise.
So the answer to this problem is simply taking $mathbf{F}(r,theta)$, compute the dot product with the derivative of the parametric curve $alpha ' (t),$ and finally compute the following integral using integration by parts $$int_0^{infty} 4 e^{-theta} sin 2theta d theta,$$ where I have used the trigonometric identity $2 sin theta cos theta = sin (2 theta).$ This will yield the correct answer, and I credit Ofek for it.
answered Aug 15 '17 at 2:39
lorenzattractorlorenzattractor
9212
answered Aug 15 '17 at 2:39
lorenzattractorlorenzattractor
9212
answered Aug 15 '17 at 2:39
lorenzattractorlorenzattractor
9212
answered Aug 15 '17 at 2:39
lorenzattractorlorenzattractor
9212
9212
add a comment |
add a comment |
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$begingroup$
Did you integrate from $0$ to $infty$? What is the correct answer according to the book?
$endgroup$
– Ofek Gillon
Mar 10 '17 at 15:33
$begingroup$
No, I haven't tried integrating from $0$ to $infty$, I'll try that right away. The book says that the answer is $frac{8}{5}.$
$endgroup$
– lorenzattractor
Mar 11 '17 at 0:08
$begingroup$
It worked! I integrated from $0$ to $infty$ and got the correct answer, but why? I really don't get it why it worked.
$endgroup$
– lorenzattractor
Mar 11 '17 at 0:48
$begingroup$
Because you need to integrate from $(1,0)$ to $(0,0)$. We see that for $theta = 0 $ we get the point $(1,0)$, but in which $theta$ do we approach $(0,0)$? You can see that you need to solve the equation $e^{-theta} = 0 Rightarrow theta= infty$ , so you need to integrate from $theta=0$ to $theta=infty$
$endgroup$
– Ofek Gillon
Mar 11 '17 at 7:50