Using integrals only, calculate the area of the greatest rectangle that can be inscribed in an ellipse.Why does using an integral to calculate an area sometimes return a negative value when using a parametric equation?What is the significance of integrating a function?Calculating area under linear lineDetermine the largest area of an ellipse enclosed by the hyperbolas ($xy=1$ and $xy=-1$)Probability of Jointly Uniform Distributed RV'sArea of the Butterfly CurveTrouble in finding the area of the curve using IntegrationArea of a Rectangle using a Double Integral in Polar CoordinatesArea under curve: integrationFind the $x$ at which the local maxima of a function ocours.
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Using integrals only, calculate the area of the greatest rectangle that can be inscribed in an ellipse.
Why does using an integral to calculate an area sometimes return a negative value when using a parametric equation?What is the significance of integrating a function?Calculating area under linear lineDetermine the largest area of an ellipse enclosed by the hyperbolas ($xy=1$ and $xy=-1$)Probability of Jointly Uniform Distributed RV'sArea of the Butterfly CurveTrouble in finding the area of the curve using IntegrationArea of a Rectangle using a Double Integral in Polar CoordinatesArea under curve: integrationFind the $x$ at which the local maxima of a function ocours.
$begingroup$
This question requires the calculation of the area with the use of integration. I know how to do this problem with the concept of maxima and minima . But when I was asked to do it with integration I assumed parametric point in the first quadrant and took the equation of the horizontal length as y=b×sin∅ and calculate the area covered by the rectangle in the first quadrant by integrating to acos∅ and multiplied it by 4.
Then I calculated the area and differentiated it and put dA/d∅ =0 from there i got ∅=π/4 as the point of maxima and finally put the value of ∅ in the equation for max area and got the value of maximum area to be $2ab$ .
Is there any other way to do this question?
Or have I done something wrong?
integration
edited Mar 22 at 8:30
dmtri
1,7692521
asked Mar 22 at 8:12
Mohit SinghMohit Singh
41
$endgroup$
add a comment |
$begingroup$
This question requires the calculation of the area with the use of integration. I know how to do this problem with the concept of maxima and minima . But when I was asked to do it with integration I assumed parametric point in the first quadrant and took the equation of the horizontal length as y=b×sin∅ and calculate the area covered by the rectangle in the first quadrant by integrating to acos∅ and multiplied it by 4.
Then I calculated the area and differentiated it and put dA/d∅ =0 from there i got ∅=π/4 as the point of maxima and finally put the value of ∅ in the equation for max area and got the value of maximum area to be $2ab$ .
Is there any other way to do this question?
Or have I done something wrong?
integration
edited Mar 22 at 8:30
dmtri
1,7692521
asked Mar 22 at 8:12
Mohit SinghMohit Singh
41
$endgroup$
$begingroup$
I think the question as written would allow for a rectangle whose sides were not "horizontal" and "vertical". "Only" integration is also a bit misleading as you are using (obvious) geometric symmetries, and also implicitly using the convexity of the ellipse. But both these observations may be overcomplicating what is intended.
$endgroup$
– Mark Bennet
Mar 22 at 8:30
add a comment |
$begingroup$
This question requires the calculation of the area with the use of integration. I know how to do this problem with the concept of maxima and minima . But when I was asked to do it with integration I assumed parametric point in the first quadrant and took the equation of the horizontal length as y=b×sin∅ and calculate the area covered by the rectangle in the first quadrant by integrating to acos∅ and multiplied it by 4.
Then I calculated the area and differentiated it and put dA/d∅ =0 from there i got ∅=π/4 as the point of maxima and finally put the value of ∅ in the equation for max area and got the value of maximum area to be $2ab$ .
Is there any other way to do this question?
Or have I done something wrong?
integration
edited Mar 22 at 8:30
dmtri
1,7692521
asked Mar 22 at 8:12
Mohit SinghMohit Singh
41
$endgroup$
This question requires the calculation of the area with the use of integration. I know how to do this problem with the concept of maxima and minima . But when I was asked to do it with integration I assumed parametric point in the first quadrant and took the equation of the horizontal length as y=b×sin∅ and calculate the area covered by the rectangle in the first quadrant by integrating to acos∅ and multiplied it by 4.
Then I calculated the area and differentiated it and put dA/d∅ =0 from there i got ∅=π/4 as the point of maxima and finally put the value of ∅ in the equation for max area and got the value of maximum area to be $2ab$ .
Is there any other way to do this question?
Or have I done something wrong?
integration
integration
edited Mar 22 at 8:30
dmtri
1,7692521
asked Mar 22 at 8:12
Mohit SinghMohit Singh
41
edited Mar 22 at 8:30
dmtri
1,7692521
asked Mar 22 at 8:12
Mohit SinghMohit Singh
41
edited Mar 22 at 8:30
dmtri
1,7692521
edited Mar 22 at 8:30
dmtri
1,7692521
edited Mar 22 at 8:30
dmtri
1,7692521
1,7692521
asked Mar 22 at 8:12
Mohit SinghMohit Singh
41
asked Mar 22 at 8:12
Mohit SinghMohit Singh
41
asked Mar 22 at 8:12
Mohit SinghMohit Singh
41
41
$begingroup$
I think the question as written would allow for a rectangle whose sides were not "horizontal" and "vertical". "Only" integration is also a bit misleading as you are using (obvious) geometric symmetries, and also implicitly using the convexity of the ellipse. But both these observations may be overcomplicating what is intended.
$endgroup$
– Mark Bennet
Mar 22 at 8:30
add a comment |
$begingroup$
I think the question as written would allow for a rectangle whose sides were not "horizontal" and "vertical". "Only" integration is also a bit misleading as you are using (obvious) geometric symmetries, and also implicitly using the convexity of the ellipse. But both these observations may be overcomplicating what is intended.
$endgroup$
– Mark Bennet
Mar 22 at 8:30
$begingroup$
I think the question as written would allow for a rectangle whose sides were not "horizontal" and "vertical". "Only" integration is also a bit misleading as you are using (obvious) geometric symmetries, and also implicitly using the convexity of the ellipse. But both these observations may be overcomplicating what is intended.
$endgroup$
– Mark Bennet
Mar 22 at 8:30
$begingroup$
I think the question as written would allow for a rectangle whose sides were not "horizontal" and "vertical". "Only" integration is also a bit misleading as you are using (obvious) geometric symmetries, and also implicitly using the convexity of the ellipse. But both these observations may be overcomplicating what is intended.
$endgroup$
– Mark Bennet
Mar 22 at 8:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For an ellipse of axes $2a$ and $2b$, $$begincasesx=acos t,\y=bsin tendcases$$ and the area of any inscribed rectangle is $$4abcos tsin t=2absin 2t.$$ The maximum is obviously $2ab$.
I can't see why/how integrals should be used here, and I can't see how this could be solved with integrals only, as you need to find the extremum.
answered Mar 22 at 8:28
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
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1 Answer
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$begingroup$
For an ellipse of axes $2a$ and $2b$, $$begincasesx=acos t,\y=bsin tendcases$$ and the area of any inscribed rectangle is $$4abcos tsin t=2absin 2t.$$ The maximum is obviously $2ab$.
I can't see why/how integrals should be used here, and I can't see how this could be solved with integrals only, as you need to find the extremum.
answered Mar 22 at 8:28
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
$begingroup$
For an ellipse of axes $2a$ and $2b$, $$begincasesx=acos t,\y=bsin tendcases$$ and the area of any inscribed rectangle is $$4abcos tsin t=2absin 2t.$$ The maximum is obviously $2ab$.
I can't see why/how integrals should be used here, and I can't see how this could be solved with integrals only, as you need to find the extremum.
answered Mar 22 at 8:28
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
$begingroup$
For an ellipse of axes $2a$ and $2b$, $$begincasesx=acos t,\y=bsin tendcases$$ and the area of any inscribed rectangle is $$4abcos tsin t=2absin 2t.$$ The maximum is obviously $2ab$.
I can't see why/how integrals should be used here, and I can't see how this could be solved with integrals only, as you need to find the extremum.
answered Mar 22 at 8:28
Yves DaoustYves Daoust
132k676230
$endgroup$
For an ellipse of axes $2a$ and $2b$, $$begincasesx=acos t,\y=bsin tendcases$$ and the area of any inscribed rectangle is $$4abcos tsin t=2absin 2t.$$ The maximum is obviously $2ab$.
I can't see why/how integrals should be used here, and I can't see how this could be solved with integrals only, as you need to find the extremum.
answered Mar 22 at 8:28
Yves DaoustYves Daoust
132k676230
edited Mar 22 at 8:35
answered Mar 22 at 8:28
Yves DaoustYves Daoust
132k676230
answered Mar 22 at 8:28
Yves DaoustYves Daoust
132k676230
answered Mar 22 at 8:28
Yves DaoustYves Daoust
132k676230
132k676230
add a comment |
add a comment |
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$begingroup$
I think the question as written would allow for a rectangle whose sides were not "horizontal" and "vertical". "Only" integration is also a bit misleading as you are using (obvious) geometric symmetries, and also implicitly using the convexity of the ellipse. But both these observations may be overcomplicating what is intended.
$endgroup$
– Mark Bennet
Mar 22 at 8:30