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What is the equation of the normal to this line?
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I am playing around with the desmos graphing calculator and I am trying to find the equation of the normal to this line where it touches the curve $x^2 + y^2 = c^2$.
$$ysin(t)+xcos(t) = c$$
where $c$ is the radius of the circle and $t$ is the angle in radians around the circle.
Here is a desmos graph I have created with the equation
I know that the normal line will have a negative reciprocal of the gradient of the tangent like in a very simple example
Example tangent
$$y = 2x$$
Normal of example tangent
$$y = -1 over 2x$$
I've tried applying this to the first equation but have had no luck. I don't think I am using the correct approach...
circle parametric tangent-line
asked Sep 9 at 0:34
Danoram
1135
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add a comment |Â
up vote
2
down vote
favorite
I am playing around with the desmos graphing calculator and I am trying to find the equation of the normal to this line where it touches the curve $x^2 + y^2 = c^2$.
$$ysin(t)+xcos(t) = c$$
where $c$ is the radius of the circle and $t$ is the angle in radians around the circle.
Here is a desmos graph I have created with the equation
I know that the normal line will have a negative reciprocal of the gradient of the tangent like in a very simple example
Example tangent
$$y = 2x$$
Normal of example tangent
$$y = -1 over 2x$$
I've tried applying this to the first equation but have had no luck. I don't think I am using the correct approach...
circle parametric tangent-line
asked Sep 9 at 0:34
Danoram
1135
New contributor
Danoram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
The normal passing through which point?
– Bernard
Sep 9 at 0:48
The point where the line touches the circle $x^2 + y^2 = c^2$. I'm not very good at explaining myself in a mathematical way sorry..
– Danoram
Sep 9 at 0:58
1
Well, the equation in your post is the equation of the tangent to the circle at this point, hence the normal is just the line through this point and the centre of the circle, and its equation is simply $y=(tan t) x$ (if $tne pmfracpi 2$).
– Bernard
Sep 9 at 1:10
Yup that is what I was looking for! As easy as it sounds I have learnt more because of it
– Danoram
Sep 9 at 1:12
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am playing around with the desmos graphing calculator and I am trying to find the equation of the normal to this line where it touches the curve $x^2 + y^2 = c^2$.
$$ysin(t)+xcos(t) = c$$
where $c$ is the radius of the circle and $t$ is the angle in radians around the circle.
Here is a desmos graph I have created with the equation
I know that the normal line will have a negative reciprocal of the gradient of the tangent like in a very simple example
Example tangent
$$y = 2x$$
Normal of example tangent
$$y = -1 over 2x$$
I've tried applying this to the first equation but have had no luck. I don't think I am using the correct approach...
circle parametric tangent-line
asked Sep 9 at 0:34
Danoram
1135
New contributor
Danoram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am playing around with the desmos graphing calculator and I am trying to find the equation of the normal to this line where it touches the curve $x^2 + y^2 = c^2$.
$$ysin(t)+xcos(t) = c$$
where $c$ is the radius of the circle and $t$ is the angle in radians around the circle.
Here is a desmos graph I have created with the equation
I know that the normal line will have a negative reciprocal of the gradient of the tangent like in a very simple example
Example tangent
$$y = 2x$$
Normal of example tangent
$$y = -1 over 2x$$
I've tried applying this to the first equation but have had no luck. I don't think I am using the correct approach...
circle parametric tangent-line
circle parametric tangent-line
asked Sep 9 at 0:34
Danoram
1135
New contributor
Danoram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Sep 9 at 0:34
Danoram
1135
New contributor
Danoram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Sep 9 at 0:59
asked Sep 9 at 0:34
Danoram
1135
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Danoram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Sep 9 at 0:34
Danoram
1135
asked Sep 9 at 0:34
Danoram
1135
1135
New contributor
Danoram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Danoram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Danoram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
The normal passing through which point?
– Bernard
Sep 9 at 0:48
The point where the line touches the circle $x^2 + y^2 = c^2$. I'm not very good at explaining myself in a mathematical way sorry..
– Danoram
Sep 9 at 0:58
1
Well, the equation in your post is the equation of the tangent to the circle at this point, hence the normal is just the line through this point and the centre of the circle, and its equation is simply $y=(tan t) x$ (if $tne pmfracpi 2$).
– Bernard
Sep 9 at 1:10
Yup that is what I was looking for! As easy as it sounds I have learnt more because of it
– Danoram
Sep 9 at 1:12
add a comment |Â
1
The normal passing through which point?
– Bernard
Sep 9 at 0:48
The point where the line touches the circle $x^2 + y^2 = c^2$. I'm not very good at explaining myself in a mathematical way sorry..
– Danoram
Sep 9 at 0:58
1
Well, the equation in your post is the equation of the tangent to the circle at this point, hence the normal is just the line through this point and the centre of the circle, and its equation is simply $y=(tan t) x$ (if $tne pmfracpi 2$).
– Bernard
Sep 9 at 1:10
Yup that is what I was looking for! As easy as it sounds I have learnt more because of it
– Danoram
Sep 9 at 1:12
1
1
The normal passing through which point?
– Bernard
Sep 9 at 0:48
The normal passing through which point?
– Bernard
Sep 9 at 0:48
The point where the line touches the circle $x^2 + y^2 = c^2$. I'm not very good at explaining myself in a mathematical way sorry..
– Danoram
Sep 9 at 0:58
The point where the line touches the circle $x^2 + y^2 = c^2$. I'm not very good at explaining myself in a mathematical way sorry..
– Danoram
Sep 9 at 0:58
1
1
Well, the equation in your post is the equation of the tangent to the circle at this point, hence the normal is just the line through this point and the centre of the circle, and its equation is simply $y=(tan t) x$ (if $tne pmfracpi 2$).
– Bernard
Sep 9 at 1:10
Well, the equation in your post is the equation of the tangent to the circle at this point, hence the normal is just the line through this point and the centre of the circle, and its equation is simply $y=(tan t) x$ (if $tne pmfracpi 2$).
– Bernard
Sep 9 at 1:10
Yup that is what I was looking for! As easy as it sounds I have learnt more because of it
– Danoram
Sep 9 at 1:12
Yup that is what I was looking for! As easy as it sounds I have learnt more because of it
– Danoram
Sep 9 at 1:12
add a comment |Â
4 Answers
4
active
oldest
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up vote
3
down vote
accepted
Well, you can rewrite the equation $ysin(t)+xcos(t)=c$ as $$y=-fraccos(t)sin(t)x+c textfor tne kpi, kinmathbbZ.$$ The slope of this line is $m_t=-fraccos(t)sin(t)$ so the equation of normal line that passes through the point $(x_0,y_0)$ must be $$y=-frac1m_t(x-x_0)+y_0implies y=tan(t)(x-x_0)+y_0 textwith tnefrackpi2.$$ The cases $t=frackpi2$ are easy.
If you want that the normal line goes through the intersection point, then you can choose $(x_0,y_0)=(0,0)$, because all the normal lines with this proprerty must contain the radius (hence $(0,0)$ must be a point of normal.) In conclusion: $y=tan(t)x$ is the equation you are looking for.
answered Sep 9 at 1:06
Ixion
691319
Thank you! of all the answers so far for me this was the easiest to understand without involving differentials etc
– Danoram
Sep 9 at 1:11
@Danoram You are welcome. :)
– Ixion
Sep 9 at 1:13
disclaimer: I appreciated all other answers too!!
– Danoram
Sep 9 at 1:15
add a comment |Â
up vote
2
down vote
We can write this as
$$
c = y sin(t) + x cos(t) = (cos(t), sin(t)) cdot (x, y) = n cdot u iff \
n cdot u = c
$$
where
$$
n=(cos(t), sin(t))
$$
is a normal vector of the line and
$$
u = (x, y)
$$
some vector from the origin to a point on the line.
In this case $n$ is a unit normal vector which makes $lvert crvert$ the distance of the line to the origin.
We see that $u=cn$ fulfills the equation of the line.
The vector
$$
hatn = (-sin(t), cos(t))
$$
is orthogonal to $n$, as
$$
hatn cdot n = (-sin(t), cos(t)) cdot (cos(t), sin(t))
= - sin(t)cos(t)+cos(t)sin(t) = 0
$$
So an equation of our normal line is
$$
hatn cdot u = d
$$
where $lvert d rvert $ is the distance of the normal line to the origin,
as $hatn$ is a unit normal vector as well.
The point where the line touches the circle $x^2+y^2=c^2$
This is the point $u = cn$. It fulfills
$$
n cdot u = ncdot(cn) = c (n cdot n) = c \
(c cos(t))^2 + (c sin(t))^2 = c^2 (cos(t)^2 + sin(t)^2) = c^2
$$
thus lies on the line and the circle.
We insert this $u$ in the equation of the normal line
$$
d
= hatn cdot u
= hatn cdot (cn)
= c (hatn cdot n)
= c , 0
= 0
$$
so we have $d=0$ and the equation
$$
hatn cdot u = 0
$$
Note: The advantage of these normal forms is that they also work for the cases $t = pmpi/2$.
answered Sep 9 at 0:49
mvw
31k22252
add a comment |Â
up vote
1
down vote
Starting with $x^2 + y^2 = c^2$ we can find the equation of the tangent line at the point (x,y) on the graph by implicitly differentiating both sides with respect to x.
$fracddx (x^2 + y^2 )=fracddx c^2$
$2x + 2yfracdydx = 0 $
$fracdydx = -fracxy$
To find the slope of the normal line, you can take the negative reciprocal of this value. If you want to find the slope of the normal line to begin with, just implicitly differentiate with respect to y instead. This will give you $fracdxdy$, which is the slope of the normal line at that point.
answered Sep 9 at 0:48
JockoCigarNab
1266
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JockoCigarNab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
1
down vote
Here's a solution with physics flavor:
You want to find the equation of the line that is normal to the circle of radius $c$ centered at origin at any time. We know the centripetal force has this property and it is in the opposite direction of the position vector. So, if $vecr(t)=(ccos(t),csin(t))$ is the position vector (i.e. the parametrization of the circle), the equation of the line is given by:
$$(x,y)= -c(cos(t),sin(t))t+c(cos(t),sin(t))$$
$$x=ccos(t)(1-t) hspace10px(1)$$
$$y=csin(t)(1-t) hspace10px(2)$$
Assuming that $cos(t)neq 0$, solve for $t$ in$(1)$ and plug it in $(2)$:
$$t=(1-fracxccos(t))$$
$$y=csin(t)big(1-(1-fracxccos(t))big)=tan(t)x$$
answered Sep 9 at 1:28
stressed out
3,6101431
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Well, you can rewrite the equation $ysin(t)+xcos(t)=c$ as $$y=-fraccos(t)sin(t)x+c textfor tne kpi, kinmathbbZ.$$ The slope of this line is $m_t=-fraccos(t)sin(t)$ so the equation of normal line that passes through the point $(x_0,y_0)$ must be $$y=-frac1m_t(x-x_0)+y_0implies y=tan(t)(x-x_0)+y_0 textwith tnefrackpi2.$$ The cases $t=frackpi2$ are easy.
If you want that the normal line goes through the intersection point, then you can choose $(x_0,y_0)=(0,0)$, because all the normal lines with this proprerty must contain the radius (hence $(0,0)$ must be a point of normal.) In conclusion: $y=tan(t)x$ is the equation you are looking for.
answered Sep 9 at 1:06
Ixion
691319
Thank you! of all the answers so far for me this was the easiest to understand without involving differentials etc
– Danoram
Sep 9 at 1:11
@Danoram You are welcome. :)
– Ixion
Sep 9 at 1:13
disclaimer: I appreciated all other answers too!!
– Danoram
Sep 9 at 1:15
add a comment |Â
up vote
3
down vote
accepted
Well, you can rewrite the equation $ysin(t)+xcos(t)=c$ as $$y=-fraccos(t)sin(t)x+c textfor tne kpi, kinmathbbZ.$$ The slope of this line is $m_t=-fraccos(t)sin(t)$ so the equation of normal line that passes through the point $(x_0,y_0)$ must be $$y=-frac1m_t(x-x_0)+y_0implies y=tan(t)(x-x_0)+y_0 textwith tnefrackpi2.$$ The cases $t=frackpi2$ are easy.
If you want that the normal line goes through the intersection point, then you can choose $(x_0,y_0)=(0,0)$, because all the normal lines with this proprerty must contain the radius (hence $(0,0)$ must be a point of normal.) In conclusion: $y=tan(t)x$ is the equation you are looking for.
answered Sep 9 at 1:06
Ixion
691319
Thank you! of all the answers so far for me this was the easiest to understand without involving differentials etc
– Danoram
Sep 9 at 1:11
@Danoram You are welcome. :)
– Ixion
Sep 9 at 1:13
disclaimer: I appreciated all other answers too!!
– Danoram
Sep 9 at 1:15
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Well, you can rewrite the equation $ysin(t)+xcos(t)=c$ as $$y=-fraccos(t)sin(t)x+c textfor tne kpi, kinmathbbZ.$$ The slope of this line is $m_t=-fraccos(t)sin(t)$ so the equation of normal line that passes through the point $(x_0,y_0)$ must be $$y=-frac1m_t(x-x_0)+y_0implies y=tan(t)(x-x_0)+y_0 textwith tnefrackpi2.$$ The cases $t=frackpi2$ are easy.
If you want that the normal line goes through the intersection point, then you can choose $(x_0,y_0)=(0,0)$, because all the normal lines with this proprerty must contain the radius (hence $(0,0)$ must be a point of normal.) In conclusion: $y=tan(t)x$ is the equation you are looking for.
answered Sep 9 at 1:06
Ixion
691319
Well, you can rewrite the equation $ysin(t)+xcos(t)=c$ as $$y=-fraccos(t)sin(t)x+c textfor tne kpi, kinmathbbZ.$$ The slope of this line is $m_t=-fraccos(t)sin(t)$ so the equation of normal line that passes through the point $(x_0,y_0)$ must be $$y=-frac1m_t(x-x_0)+y_0implies y=tan(t)(x-x_0)+y_0 textwith tnefrackpi2.$$ The cases $t=frackpi2$ are easy.
If you want that the normal line goes through the intersection point, then you can choose $(x_0,y_0)=(0,0)$, because all the normal lines with this proprerty must contain the radius (hence $(0,0)$ must be a point of normal.) In conclusion: $y=tan(t)x$ is the equation you are looking for.
answered Sep 9 at 1:06
Ixion
691319
answered Sep 9 at 1:06
Ixion
691319
answered Sep 9 at 1:06
Ixion
691319
answered Sep 9 at 1:06
Ixion
691319
691319
Thank you! of all the answers so far for me this was the easiest to understand without involving differentials etc
– Danoram
Sep 9 at 1:11
@Danoram You are welcome. :)
– Ixion
Sep 9 at 1:13
disclaimer: I appreciated all other answers too!!
– Danoram
Sep 9 at 1:15
add a comment |Â
Thank you! of all the answers so far for me this was the easiest to understand without involving differentials etc
– Danoram
Sep 9 at 1:11
@Danoram You are welcome. :)
– Ixion
Sep 9 at 1:13
disclaimer: I appreciated all other answers too!!
– Danoram
Sep 9 at 1:15
Thank you! of all the answers so far for me this was the easiest to understand without involving differentials etc
– Danoram
Sep 9 at 1:11
Thank you! of all the answers so far for me this was the easiest to understand without involving differentials etc
– Danoram
Sep 9 at 1:11
@Danoram You are welcome. :)
– Ixion
Sep 9 at 1:13
@Danoram You are welcome. :)
– Ixion
Sep 9 at 1:13
disclaimer: I appreciated all other answers too!!
– Danoram
Sep 9 at 1:15
disclaimer: I appreciated all other answers too!!
– Danoram
Sep 9 at 1:15
add a comment |Â
up vote
2
down vote
We can write this as
$$
c = y sin(t) + x cos(t) = (cos(t), sin(t)) cdot (x, y) = n cdot u iff \
n cdot u = c
$$
where
$$
n=(cos(t), sin(t))
$$
is a normal vector of the line and
$$
u = (x, y)
$$
some vector from the origin to a point on the line.
In this case $n$ is a unit normal vector which makes $lvert crvert$ the distance of the line to the origin.
We see that $u=cn$ fulfills the equation of the line.
The vector
$$
hatn = (-sin(t), cos(t))
$$
is orthogonal to $n$, as
$$
hatn cdot n = (-sin(t), cos(t)) cdot (cos(t), sin(t))
= - sin(t)cos(t)+cos(t)sin(t) = 0
$$
So an equation of our normal line is
$$
hatn cdot u = d
$$
where $lvert d rvert $ is the distance of the normal line to the origin,
as $hatn$ is a unit normal vector as well.
The point where the line touches the circle $x^2+y^2=c^2$
This is the point $u = cn$. It fulfills
$$
n cdot u = ncdot(cn) = c (n cdot n) = c \
(c cos(t))^2 + (c sin(t))^2 = c^2 (cos(t)^2 + sin(t)^2) = c^2
$$
thus lies on the line and the circle.
We insert this $u$ in the equation of the normal line
$$
d
= hatn cdot u
= hatn cdot (cn)
= c (hatn cdot n)
= c , 0
= 0
$$
so we have $d=0$ and the equation
$$
hatn cdot u = 0
$$
Note: The advantage of these normal forms is that they also work for the cases $t = pmpi/2$.
answered Sep 9 at 0:49
mvw
31k22252
add a comment |Â
up vote
2
down vote
We can write this as
$$
c = y sin(t) + x cos(t) = (cos(t), sin(t)) cdot (x, y) = n cdot u iff \
n cdot u = c
$$
where
$$
n=(cos(t), sin(t))
$$
is a normal vector of the line and
$$
u = (x, y)
$$
some vector from the origin to a point on the line.
In this case $n$ is a unit normal vector which makes $lvert crvert$ the distance of the line to the origin.
We see that $u=cn$ fulfills the equation of the line.
The vector
$$
hatn = (-sin(t), cos(t))
$$
is orthogonal to $n$, as
$$
hatn cdot n = (-sin(t), cos(t)) cdot (cos(t), sin(t))
= - sin(t)cos(t)+cos(t)sin(t) = 0
$$
So an equation of our normal line is
$$
hatn cdot u = d
$$
where $lvert d rvert $ is the distance of the normal line to the origin,
as $hatn$ is a unit normal vector as well.
The point where the line touches the circle $x^2+y^2=c^2$
This is the point $u = cn$. It fulfills
$$
n cdot u = ncdot(cn) = c (n cdot n) = c \
(c cos(t))^2 + (c sin(t))^2 = c^2 (cos(t)^2 + sin(t)^2) = c^2
$$
thus lies on the line and the circle.
We insert this $u$ in the equation of the normal line
$$
d
= hatn cdot u
= hatn cdot (cn)
= c (hatn cdot n)
= c , 0
= 0
$$
so we have $d=0$ and the equation
$$
hatn cdot u = 0
$$
Note: The advantage of these normal forms is that they also work for the cases $t = pmpi/2$.
answered Sep 9 at 0:49
mvw
31k22252
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We can write this as
$$
c = y sin(t) + x cos(t) = (cos(t), sin(t)) cdot (x, y) = n cdot u iff \
n cdot u = c
$$
where
$$
n=(cos(t), sin(t))
$$
is a normal vector of the line and
$$
u = (x, y)
$$
some vector from the origin to a point on the line.
In this case $n$ is a unit normal vector which makes $lvert crvert$ the distance of the line to the origin.
We see that $u=cn$ fulfills the equation of the line.
The vector
$$
hatn = (-sin(t), cos(t))
$$
is orthogonal to $n$, as
$$
hatn cdot n = (-sin(t), cos(t)) cdot (cos(t), sin(t))
= - sin(t)cos(t)+cos(t)sin(t) = 0
$$
So an equation of our normal line is
$$
hatn cdot u = d
$$
where $lvert d rvert $ is the distance of the normal line to the origin,
as $hatn$ is a unit normal vector as well.
The point where the line touches the circle $x^2+y^2=c^2$
This is the point $u = cn$. It fulfills
$$
n cdot u = ncdot(cn) = c (n cdot n) = c \
(c cos(t))^2 + (c sin(t))^2 = c^2 (cos(t)^2 + sin(t)^2) = c^2
$$
thus lies on the line and the circle.
We insert this $u$ in the equation of the normal line
$$
d
= hatn cdot u
= hatn cdot (cn)
= c (hatn cdot n)
= c , 0
= 0
$$
so we have $d=0$ and the equation
$$
hatn cdot u = 0
$$
Note: The advantage of these normal forms is that they also work for the cases $t = pmpi/2$.
answered Sep 9 at 0:49
mvw
31k22252
We can write this as
$$
c = y sin(t) + x cos(t) = (cos(t), sin(t)) cdot (x, y) = n cdot u iff \
n cdot u = c
$$
where
$$
n=(cos(t), sin(t))
$$
is a normal vector of the line and
$$
u = (x, y)
$$
some vector from the origin to a point on the line.
In this case $n$ is a unit normal vector which makes $lvert crvert$ the distance of the line to the origin.
We see that $u=cn$ fulfills the equation of the line.
The vector
$$
hatn = (-sin(t), cos(t))
$$
is orthogonal to $n$, as
$$
hatn cdot n = (-sin(t), cos(t)) cdot (cos(t), sin(t))
= - sin(t)cos(t)+cos(t)sin(t) = 0
$$
So an equation of our normal line is
$$
hatn cdot u = d
$$
where $lvert d rvert $ is the distance of the normal line to the origin,
as $hatn$ is a unit normal vector as well.
The point where the line touches the circle $x^2+y^2=c^2$
This is the point $u = cn$. It fulfills
$$
n cdot u = ncdot(cn) = c (n cdot n) = c \
(c cos(t))^2 + (c sin(t))^2 = c^2 (cos(t)^2 + sin(t)^2) = c^2
$$
thus lies on the line and the circle.
We insert this $u$ in the equation of the normal line
$$
d
= hatn cdot u
= hatn cdot (cn)
= c (hatn cdot n)
= c , 0
= 0
$$
so we have $d=0$ and the equation
$$
hatn cdot u = 0
$$
Note: The advantage of these normal forms is that they also work for the cases $t = pmpi/2$.
answered Sep 9 at 0:49
mvw
31k22252
edited Sep 9 at 1:34
answered Sep 9 at 0:49
mvw
31k22252
answered Sep 9 at 0:49
mvw
31k22252
answered Sep 9 at 0:49
mvw
31k22252
31k22252
add a comment |Â
add a comment |Â
up vote
1
down vote
Starting with $x^2 + y^2 = c^2$ we can find the equation of the tangent line at the point (x,y) on the graph by implicitly differentiating both sides with respect to x.
$fracddx (x^2 + y^2 )=fracddx c^2$
$2x + 2yfracdydx = 0 $
$fracdydx = -fracxy$
To find the slope of the normal line, you can take the negative reciprocal of this value. If you want to find the slope of the normal line to begin with, just implicitly differentiate with respect to y instead. This will give you $fracdxdy$, which is the slope of the normal line at that point.
answered Sep 9 at 0:48
JockoCigarNab
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Starting with $x^2 + y^2 = c^2$ we can find the equation of the tangent line at the point (x,y) on the graph by implicitly differentiating both sides with respect to x.
$fracddx (x^2 + y^2 )=fracddx c^2$
$2x + 2yfracdydx = 0 $
$fracdydx = -fracxy$
To find the slope of the normal line, you can take the negative reciprocal of this value. If you want to find the slope of the normal line to begin with, just implicitly differentiate with respect to y instead. This will give you $fracdxdy$, which is the slope of the normal line at that point.
answered Sep 9 at 0:48
JockoCigarNab
1266
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up vote
1
down vote
up vote
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Starting with $x^2 + y^2 = c^2$ we can find the equation of the tangent line at the point (x,y) on the graph by implicitly differentiating both sides with respect to x.
$fracddx (x^2 + y^2 )=fracddx c^2$
$2x + 2yfracdydx = 0 $
$fracdydx = -fracxy$
To find the slope of the normal line, you can take the negative reciprocal of this value. If you want to find the slope of the normal line to begin with, just implicitly differentiate with respect to y instead. This will give you $fracdxdy$, which is the slope of the normal line at that point.
answered Sep 9 at 0:48
JockoCigarNab
1266
New contributor
JockoCigarNab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Starting with $x^2 + y^2 = c^2$ we can find the equation of the tangent line at the point (x,y) on the graph by implicitly differentiating both sides with respect to x.
$fracddx (x^2 + y^2 )=fracddx c^2$
$2x + 2yfracdydx = 0 $
$fracdydx = -fracxy$
To find the slope of the normal line, you can take the negative reciprocal of this value. If you want to find the slope of the normal line to begin with, just implicitly differentiate with respect to y instead. This will give you $fracdxdy$, which is the slope of the normal line at that point.
answered Sep 9 at 0:48
JockoCigarNab
1266
New contributor
JockoCigarNab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered Sep 9 at 0:48
JockoCigarNab
1266
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answered Sep 9 at 0:48
JockoCigarNab
1266
answered Sep 9 at 0:48
JockoCigarNab
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1266
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Here's a solution with physics flavor:
You want to find the equation of the line that is normal to the circle of radius $c$ centered at origin at any time. We know the centripetal force has this property and it is in the opposite direction of the position vector. So, if $vecr(t)=(ccos(t),csin(t))$ is the position vector (i.e. the parametrization of the circle), the equation of the line is given by:
$$(x,y)= -c(cos(t),sin(t))t+c(cos(t),sin(t))$$
$$x=ccos(t)(1-t) hspace10px(1)$$
$$y=csin(t)(1-t) hspace10px(2)$$
Assuming that $cos(t)neq 0$, solve for $t$ in$(1)$ and plug it in $(2)$:
$$t=(1-fracxccos(t))$$
$$y=csin(t)big(1-(1-fracxccos(t))big)=tan(t)x$$
answered Sep 9 at 1:28
stressed out
3,6101431
add a comment |Â
up vote
1
down vote
Here's a solution with physics flavor:
You want to find the equation of the line that is normal to the circle of radius $c$ centered at origin at any time. We know the centripetal force has this property and it is in the opposite direction of the position vector. So, if $vecr(t)=(ccos(t),csin(t))$ is the position vector (i.e. the parametrization of the circle), the equation of the line is given by:
$$(x,y)= -c(cos(t),sin(t))t+c(cos(t),sin(t))$$
$$x=ccos(t)(1-t) hspace10px(1)$$
$$y=csin(t)(1-t) hspace10px(2)$$
Assuming that $cos(t)neq 0$, solve for $t$ in$(1)$ and plug it in $(2)$:
$$t=(1-fracxccos(t))$$
$$y=csin(t)big(1-(1-fracxccos(t))big)=tan(t)x$$
answered Sep 9 at 1:28
stressed out
3,6101431
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here's a solution with physics flavor:
You want to find the equation of the line that is normal to the circle of radius $c$ centered at origin at any time. We know the centripetal force has this property and it is in the opposite direction of the position vector. So, if $vecr(t)=(ccos(t),csin(t))$ is the position vector (i.e. the parametrization of the circle), the equation of the line is given by:
$$(x,y)= -c(cos(t),sin(t))t+c(cos(t),sin(t))$$
$$x=ccos(t)(1-t) hspace10px(1)$$
$$y=csin(t)(1-t) hspace10px(2)$$
Assuming that $cos(t)neq 0$, solve for $t$ in$(1)$ and plug it in $(2)$:
$$t=(1-fracxccos(t))$$
$$y=csin(t)big(1-(1-fracxccos(t))big)=tan(t)x$$
answered Sep 9 at 1:28
stressed out
3,6101431
Here's a solution with physics flavor:
You want to find the equation of the line that is normal to the circle of radius $c$ centered at origin at any time. We know the centripetal force has this property and it is in the opposite direction of the position vector. So, if $vecr(t)=(ccos(t),csin(t))$ is the position vector (i.e. the parametrization of the circle), the equation of the line is given by:
$$(x,y)= -c(cos(t),sin(t))t+c(cos(t),sin(t))$$
$$x=ccos(t)(1-t) hspace10px(1)$$
$$y=csin(t)(1-t) hspace10px(2)$$
Assuming that $cos(t)neq 0$, solve for $t$ in$(1)$ and plug it in $(2)$:
$$t=(1-fracxccos(t))$$
$$y=csin(t)big(1-(1-fracxccos(t))big)=tan(t)x$$
answered Sep 9 at 1:28
stressed out
3,6101431
answered Sep 9 at 1:28
stressed out
3,6101431
answered Sep 9 at 1:28
stressed out
3,6101431
answered Sep 9 at 1:28
stressed out
3,6101431
3,6101431
add a comment |Â
add a comment |Â
Danoram is a new contributor. Be nice, and check out our Code of Conduct.
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1
The normal passing through which point?
– Bernard
Sep 9 at 0:48
The point where the line touches the circle $x^2 + y^2 = c^2$. I'm not very good at explaining myself in a mathematical way sorry..
– Danoram
Sep 9 at 0:58
1
Well, the equation in your post is the equation of the tangent to the circle at this point, hence the normal is just the line through this point and the centre of the circle, and its equation is simply $y=(tan t) x$ (if $tne pmfracpi 2$).
– Bernard
Sep 9 at 1:10
Yup that is what I was looking for! As easy as it sounds I have learnt more because of it
– Danoram
Sep 9 at 1:12