Parametric Equations for semicircle
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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
asked Jan 2 at 17:57
MwaxMwax
458
$endgroup$
add a comment |
$begingroup$
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
asked Jan 2 at 17:57
MwaxMwax
458
$endgroup$
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
add a comment |
$begingroup$
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
asked Jan 2 at 17:57
MwaxMwax
458
$endgroup$
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
asked Jan 2 at 17:57
MwaxMwax
458
asked Jan 2 at 17:57
MwaxMwax
458
edited Jan 2 at 18:10
Mwax
asked Jan 2 at 17:57
MwaxMwax
458
asked Jan 2 at 17:57
MwaxMwax
458
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 2 at 18:04
Ben WBen W
2,249615
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
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$begingroup$
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.
answered Jan 2 at 18:04
Ben WBen W
2,249615
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 2 at 18:04
Ben WBen W
2,249615
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 2 at 18:04
Ben WBen W
2,249615
$endgroup$
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.
answered Jan 2 at 18:04
Ben WBen W
2,249615
answered Jan 2 at 18:04
Ben WBen W
2,249615
answered Jan 2 at 18:04
Ben WBen W
2,249615
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
add a comment |
$begingroup$
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
add a comment |
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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