Parametric Equations for semicircle












2












$begingroup$


How does one find the parametric equation for the semicircle $x^2 + y^2 = a^2$, $y > 0$, using as parameter the slope $t = frac{dy}{dx}$ of the tangent to the curve at $(x, y)$?



The solution given is $x = frac{-at}{sqrt{1 + t^2}}$, $y = {a}{sqrt{1 + t^2}}$, $-infty < t < infty$. But I cannot see how this solution has been obtained.



This is what I have so far. Use implicit differentiation to get $frac{dy}{dx} =frac{- x}{y}$. But where does one go from there?






calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Shouldn’t it be $frac{dy}{dx}$?
    $endgroup$
    – Anurag A
    Jan 2 at 17:58












  • $begingroup$
    yes! It should be that.
    $endgroup$
    – Mwax
    Jan 2 at 17:59






  • 1




    $begingroup$
    This has the tone of a command and not a question.
    $endgroup$
    – Mason
    Jan 2 at 18:00










  • $begingroup$
    Duly noted and rephrased. Thank you for pointing that out.
    $endgroup$
    – Mwax
    Jan 2 at 18:15
















2












$begingroup$


How does one find the parametric equation for the semicircle $x^2 + y^2 = a^2$, $y > 0$, using as parameter the slope $t = frac{dy}{dx}$ of the tangent to the curve at $(x, y)$?



The solution given is $x = frac{-at}{sqrt{1 + t^2}}$, $y = {a}{sqrt{1 + t^2}}$, $-infty < t < infty$. But I cannot see how this solution has been obtained.



This is what I have so far. Use implicit differentiation to get $frac{dy}{dx} =frac{- x}{y}$. But where does one go from there?






calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Shouldn’t it be $frac{dy}{dx}$?
    $endgroup$
    – Anurag A
    Jan 2 at 17:58












  • $begingroup$
    yes! It should be that.
    $endgroup$
    – Mwax
    Jan 2 at 17:59






  • 1




    $begingroup$
    This has the tone of a command and not a question.
    $endgroup$
    – Mason
    Jan 2 at 18:00










  • $begingroup$
    Duly noted and rephrased. Thank you for pointing that out.
    $endgroup$
    – Mwax
    Jan 2 at 18:15














2












2








2





$begingroup$


How does one find the parametric equation for the semicircle $x^2 + y^2 = a^2$, $y > 0$, using as parameter the slope $t = frac{dy}{dx}$ of the tangent to the curve at $(x, y)$?



The solution given is $x = frac{-at}{sqrt{1 + t^2}}$, $y = {a}{sqrt{1 + t^2}}$, $-infty < t < infty$. But I cannot see how this solution has been obtained.



This is what I have so far. Use implicit differentiation to get $frac{dy}{dx} =frac{- x}{y}$. But where does one go from there?






calculus







share|cite|improve this question











$endgroup$




How does one find the parametric equation for the semicircle $x^2 + y^2 = a^2$, $y > 0$, using as parameter the slope $t = frac{dy}{dx}$ of the tangent to the curve at $(x, y)$?



The solution given is $x = frac{-at}{sqrt{1 + t^2}}$, $y = {a}{sqrt{1 + t^2}}$, $-infty < t < infty$. But I cannot see how this solution has been obtained.



This is what I have so far. Use implicit differentiation to get $frac{dy}{dx} =frac{- x}{y}$. But where does one go from there?






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 18:10







Mwax

















asked Jan 2 at 17:57









MwaxMwax

458




458








  • 1




    $begingroup$
    Shouldn’t it be $frac{dy}{dx}$?
    $endgroup$
    – Anurag A
    Jan 2 at 17:58












  • $begingroup$
    yes! It should be that.
    $endgroup$
    – Mwax
    Jan 2 at 17:59






  • 1




    $begingroup$
    This has the tone of a command and not a question.
    $endgroup$
    – Mason
    Jan 2 at 18:00










  • $begingroup$
    Duly noted and rephrased. Thank you for pointing that out.
    $endgroup$
    – Mwax
    Jan 2 at 18:15














  • 1




    $begingroup$
    Shouldn’t it be $frac{dy}{dx}$?
    $endgroup$
    – Anurag A
    Jan 2 at 17:58












  • $begingroup$
    yes! It should be that.
    $endgroup$
    – Mwax
    Jan 2 at 17:59






  • 1




    $begingroup$
    This has the tone of a command and not a question.
    $endgroup$
    – Mason
    Jan 2 at 18:00










  • $begingroup$
    Duly noted and rephrased. Thank you for pointing that out.
    $endgroup$
    – Mwax
    Jan 2 at 18:15








1




1




$begingroup$
Shouldn’t it be $frac{dy}{dx}$?
$endgroup$
– Anurag A
Jan 2 at 17:58






$begingroup$
Shouldn’t it be $frac{dy}{dx}$?
$endgroup$
– Anurag A
Jan 2 at 17:58














$begingroup$
yes! It should be that.
$endgroup$
– Mwax
Jan 2 at 17:59




$begingroup$
yes! It should be that.
$endgroup$
– Mwax
Jan 2 at 17:59




1




1




$begingroup$
This has the tone of a command and not a question.
$endgroup$
– Mason
Jan 2 at 18:00




$begingroup$
This has the tone of a command and not a question.
$endgroup$
– Mason
Jan 2 at 18:00












$begingroup$
Duly noted and rephrased. Thank you for pointing that out.
$endgroup$
– Mwax
Jan 2 at 18:15




$begingroup$
Duly noted and rephrased. Thank you for pointing that out.
$endgroup$
– Mwax
Jan 2 at 18:15










1 Answer
1






active

oldest

votes


















3












$begingroup$

Use implicit differentiation. You get $2x+2yfrac{dy}{dx}=0$. Replace $frac{dy}{dx}$ with $t$ and then solve for $x$. Next, plug in the solution to $x^2+y^2=a^2$. Solve the resulting equation for $y$ and plug that back into your solution for $x$. Boom, done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! How did you know that you can just replace $frac{dy}{dx}$ with t?
    $endgroup$
    – Mwax
    Jan 2 at 18:14










  • $begingroup$
    In your question statement you demand that $t=frac{dy}{dx}$.
    $endgroup$
    – Ben W
    Jan 2 at 18:14










  • $begingroup$
    I do demand that. Why couldn't we just plug in the value of $frac{dy}{dx}$ into the usual parametric equations of a circle to get something like this: $x = a cos (frac{dy}{dx})$ and $y = a sin(frac{dy}{dx})$ for $0 leq t leq pi$?
    $endgroup$
    – Mwax
    Jan 2 at 18:18








  • 1




    $begingroup$
    I'm not sure I know what you mean. The "usual" parametric equations of a circle are $x=acos(theta),y=asin(theta)$. By the usual polar conversion formula we have that $tan(theta)=y/x$. The slope $t$ of the tangent line is perpendicular to the line thru the origin with angle $theta$, and thus satisfies $tan(theta)=-1/t$. Now $theta=text{arctan}(-1/t)$ which means $x=acos(text{arctan}(-1/t))$ and $y=asin(text{arctan}(-1/t))$, maybe with a $pm$ decision to be made. But such formulas are rather nasty and should be simplified. At this point we are doing more work than necessary.
    $endgroup$
    – Ben W
    Jan 2 at 18:26













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1 Answer
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oldest

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1 Answer
1






active

oldest

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active

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active

oldest

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3












$begingroup$

Use implicit differentiation. You get $2x+2yfrac{dy}{dx}=0$. Replace $frac{dy}{dx}$ with $t$ and then solve for $x$. Next, plug in the solution to $x^2+y^2=a^2$. Solve the resulting equation for $y$ and plug that back into your solution for $x$. Boom, done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! How did you know that you can just replace $frac{dy}{dx}$ with t?
    $endgroup$
    – Mwax
    Jan 2 at 18:14










  • $begingroup$
    In your question statement you demand that $t=frac{dy}{dx}$.
    $endgroup$
    – Ben W
    Jan 2 at 18:14










  • $begingroup$
    I do demand that. Why couldn't we just plug in the value of $frac{dy}{dx}$ into the usual parametric equations of a circle to get something like this: $x = a cos (frac{dy}{dx})$ and $y = a sin(frac{dy}{dx})$ for $0 leq t leq pi$?
    $endgroup$
    – Mwax
    Jan 2 at 18:18








  • 1




    $begingroup$
    I'm not sure I know what you mean. The "usual" parametric equations of a circle are $x=acos(theta),y=asin(theta)$. By the usual polar conversion formula we have that $tan(theta)=y/x$. The slope $t$ of the tangent line is perpendicular to the line thru the origin with angle $theta$, and thus satisfies $tan(theta)=-1/t$. Now $theta=text{arctan}(-1/t)$ which means $x=acos(text{arctan}(-1/t))$ and $y=asin(text{arctan}(-1/t))$, maybe with a $pm$ decision to be made. But such formulas are rather nasty and should be simplified. At this point we are doing more work than necessary.
    $endgroup$
    – Ben W
    Jan 2 at 18:26


















3












$begingroup$

Use implicit differentiation. You get $2x+2yfrac{dy}{dx}=0$. Replace $frac{dy}{dx}$ with $t$ and then solve for $x$. Next, plug in the solution to $x^2+y^2=a^2$. Solve the resulting equation for $y$ and plug that back into your solution for $x$. Boom, done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! How did you know that you can just replace $frac{dy}{dx}$ with t?
    $endgroup$
    – Mwax
    Jan 2 at 18:14










  • $begingroup$
    In your question statement you demand that $t=frac{dy}{dx}$.
    $endgroup$
    – Ben W
    Jan 2 at 18:14










  • $begingroup$
    I do demand that. Why couldn't we just plug in the value of $frac{dy}{dx}$ into the usual parametric equations of a circle to get something like this: $x = a cos (frac{dy}{dx})$ and $y = a sin(frac{dy}{dx})$ for $0 leq t leq pi$?
    $endgroup$
    – Mwax
    Jan 2 at 18:18








  • 1




    $begingroup$
    I'm not sure I know what you mean. The "usual" parametric equations of a circle are $x=acos(theta),y=asin(theta)$. By the usual polar conversion formula we have that $tan(theta)=y/x$. The slope $t$ of the tangent line is perpendicular to the line thru the origin with angle $theta$, and thus satisfies $tan(theta)=-1/t$. Now $theta=text{arctan}(-1/t)$ which means $x=acos(text{arctan}(-1/t))$ and $y=asin(text{arctan}(-1/t))$, maybe with a $pm$ decision to be made. But such formulas are rather nasty and should be simplified. At this point we are doing more work than necessary.
    $endgroup$
    – Ben W
    Jan 2 at 18:26
















3












3








3





$begingroup$

Use implicit differentiation. You get $2x+2yfrac{dy}{dx}=0$. Replace $frac{dy}{dx}$ with $t$ and then solve for $x$. Next, plug in the solution to $x^2+y^2=a^2$. Solve the resulting equation for $y$ and plug that back into your solution for $x$. Boom, done.






share|cite|improve this answer









$endgroup$



Use implicit differentiation. You get $2x+2yfrac{dy}{dx}=0$. Replace $frac{dy}{dx}$ with $t$ and then solve for $x$. Next, plug in the solution to $x^2+y^2=a^2$. Solve the resulting equation for $y$ and plug that back into your solution for $x$. Boom, done.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 2 at 18:04









Ben WBen W

2,249615




2,249615












  • $begingroup$
    Thank you! How did you know that you can just replace $frac{dy}{dx}$ with t?
    $endgroup$
    – Mwax
    Jan 2 at 18:14










  • $begingroup$
    In your question statement you demand that $t=frac{dy}{dx}$.
    $endgroup$
    – Ben W
    Jan 2 at 18:14










  • $begingroup$
    I do demand that. Why couldn't we just plug in the value of $frac{dy}{dx}$ into the usual parametric equations of a circle to get something like this: $x = a cos (frac{dy}{dx})$ and $y = a sin(frac{dy}{dx})$ for $0 leq t leq pi$?
    $endgroup$
    – Mwax
    Jan 2 at 18:18








  • 1




    $begingroup$
    I'm not sure I know what you mean. The "usual" parametric equations of a circle are $x=acos(theta),y=asin(theta)$. By the usual polar conversion formula we have that $tan(theta)=y/x$. The slope $t$ of the tangent line is perpendicular to the line thru the origin with angle $theta$, and thus satisfies $tan(theta)=-1/t$. Now $theta=text{arctan}(-1/t)$ which means $x=acos(text{arctan}(-1/t))$ and $y=asin(text{arctan}(-1/t))$, maybe with a $pm$ decision to be made. But such formulas are rather nasty and should be simplified. At this point we are doing more work than necessary.
    $endgroup$
    – Ben W
    Jan 2 at 18:26




















  • $begingroup$
    Thank you! How did you know that you can just replace $frac{dy}{dx}$ with t?
    $endgroup$
    – Mwax
    Jan 2 at 18:14










  • $begingroup$
    In your question statement you demand that $t=frac{dy}{dx}$.
    $endgroup$
    – Ben W
    Jan 2 at 18:14










  • $begingroup$
    I do demand that. Why couldn't we just plug in the value of $frac{dy}{dx}$ into the usual parametric equations of a circle to get something like this: $x = a cos (frac{dy}{dx})$ and $y = a sin(frac{dy}{dx})$ for $0 leq t leq pi$?
    $endgroup$
    – Mwax
    Jan 2 at 18:18








  • 1




    $begingroup$
    I'm not sure I know what you mean. The "usual" parametric equations of a circle are $x=acos(theta),y=asin(theta)$. By the usual polar conversion formula we have that $tan(theta)=y/x$. The slope $t$ of the tangent line is perpendicular to the line thru the origin with angle $theta$, and thus satisfies $tan(theta)=-1/t$. Now $theta=text{arctan}(-1/t)$ which means $x=acos(text{arctan}(-1/t))$ and $y=asin(text{arctan}(-1/t))$, maybe with a $pm$ decision to be made. But such formulas are rather nasty and should be simplified. At this point we are doing more work than necessary.
    $endgroup$
    – Ben W
    Jan 2 at 18:26


















$begingroup$
Thank you! How did you know that you can just replace $frac{dy}{dx}$ with t?
$endgroup$
– Mwax
Jan 2 at 18:14




$begingroup$
Thank you! How did you know that you can just replace $frac{dy}{dx}$ with t?
$endgroup$
– Mwax
Jan 2 at 18:14












$begingroup$
In your question statement you demand that $t=frac{dy}{dx}$.
$endgroup$
– Ben W
Jan 2 at 18:14




$begingroup$
In your question statement you demand that $t=frac{dy}{dx}$.
$endgroup$
– Ben W
Jan 2 at 18:14












$begingroup$
I do demand that. Why couldn't we just plug in the value of $frac{dy}{dx}$ into the usual parametric equations of a circle to get something like this: $x = a cos (frac{dy}{dx})$ and $y = a sin(frac{dy}{dx})$ for $0 leq t leq pi$?
$endgroup$
– Mwax
Jan 2 at 18:18






$begingroup$
I do demand that. Why couldn't we just plug in the value of $frac{dy}{dx}$ into the usual parametric equations of a circle to get something like this: $x = a cos (frac{dy}{dx})$ and $y = a sin(frac{dy}{dx})$ for $0 leq t leq pi$?
$endgroup$
– Mwax
Jan 2 at 18:18






1




1




$begingroup$
I'm not sure I know what you mean. The "usual" parametric equations of a circle are $x=acos(theta),y=asin(theta)$. By the usual polar conversion formula we have that $tan(theta)=y/x$. The slope $t$ of the tangent line is perpendicular to the line thru the origin with angle $theta$, and thus satisfies $tan(theta)=-1/t$. Now $theta=text{arctan}(-1/t)$ which means $x=acos(text{arctan}(-1/t))$ and $y=asin(text{arctan}(-1/t))$, maybe with a $pm$ decision to be made. But such formulas are rather nasty and should be simplified. At this point we are doing more work than necessary.
$endgroup$
– Ben W
Jan 2 at 18:26






$begingroup$
I'm not sure I know what you mean. The "usual" parametric equations of a circle are $x=acos(theta),y=asin(theta)$. By the usual polar conversion formula we have that $tan(theta)=y/x$. The slope $t$ of the tangent line is perpendicular to the line thru the origin with angle $theta$, and thus satisfies $tan(theta)=-1/t$. Now $theta=text{arctan}(-1/t)$ which means $x=acos(text{arctan}(-1/t))$ and $y=asin(text{arctan}(-1/t))$, maybe with a $pm$ decision to be made. But such formulas are rather nasty and should be simplified. At this point we are doing more work than necessary.
$endgroup$
– Ben W
Jan 2 at 18:26




















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