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How to find a unit normal vector to the given curve?
$begingroup$
Find the unit normal vector to the curve $r(t) = (3sin t)i + (3cos t)j + 4t k$ at point $( pi /2, 0,1)$.
The tangent vector of this curve is $(3 cos t)i -(3 sin t)j + 4k$ and unit normal vector should be perpendicular to this vector at given point.
But I couldn't get the final answer. How to find unit normal vector at the given point$?$
vector-analysis
asked Jan 30 at 20:30
MathsaddictMathsaddict
3669
$endgroup$
add a comment |
$begingroup$
Find the unit normal vector to the curve $r(t) = (3sin t)i + (3cos t)j + 4t k$ at point $( pi /2, 0,1)$.
The tangent vector of this curve is $(3 cos t)i -(3 sin t)j + 4k$ and unit normal vector should be perpendicular to this vector at given point.
But I couldn't get the final answer. How to find unit normal vector at the given point$?$
vector-analysis
asked Jan 30 at 20:30
MathsaddictMathsaddict
3669
$endgroup$
1
$begingroup$
This point is not on the curve?
$endgroup$
– Peter Foreman
Jan 30 at 20:39
$begingroup$
This point is on the curve. I couldn't find any value of $t$ for this point
$endgroup$
– Mathsaddict
Jan 30 at 20:43
$begingroup$
Therefore it is not on the curve?
$endgroup$
– Peter Foreman
Jan 30 at 20:45
$begingroup$
So, there's no solution?
$endgroup$
– Mathsaddict
Jan 30 at 20:46
1
$begingroup$
If the point is on the curve one can find the value of t as 1/4 of the k coefficient. Then use the formula you mentioned for a correct answer. But this point is not on the curve, so the normal to the curve cannot be found.
$endgroup$
– Peter Foreman
Jan 30 at 20:47
add a comment |
$begingroup$
Find the unit normal vector to the curve $r(t) = (3sin t)i + (3cos t)j + 4t k$ at point $( pi /2, 0,1)$.
The tangent vector of this curve is $(3 cos t)i -(3 sin t)j + 4k$ and unit normal vector should be perpendicular to this vector at given point.
But I couldn't get the final answer. How to find unit normal vector at the given point$?$
vector-analysis
asked Jan 30 at 20:30
MathsaddictMathsaddict
3669
$endgroup$
Find the unit normal vector to the curve $r(t) = (3sin t)i + (3cos t)j + 4t k$ at point $( pi /2, 0,1)$.
The tangent vector of this curve is $(3 cos t)i -(3 sin t)j + 4k$ and unit normal vector should be perpendicular to this vector at given point.
But I couldn't get the final answer. How to find unit normal vector at the given point$?$
vector-analysis
vector-analysis
asked Jan 30 at 20:30
MathsaddictMathsaddict
3669
asked Jan 30 at 20:30
MathsaddictMathsaddict
3669
asked Jan 30 at 20:30
MathsaddictMathsaddict
3669
asked Jan 30 at 20:30
MathsaddictMathsaddict
3669
asked Jan 30 at 20:30
MathsaddictMathsaddict
3669
3669
1
$begingroup$
This point is not on the curve?
$endgroup$
– Peter Foreman
Jan 30 at 20:39
$begingroup$
This point is on the curve. I couldn't find any value of $t$ for this point
$endgroup$
– Mathsaddict
Jan 30 at 20:43
$begingroup$
Therefore it is not on the curve?
$endgroup$
– Peter Foreman
Jan 30 at 20:45
$begingroup$
So, there's no solution?
$endgroup$
– Mathsaddict
Jan 30 at 20:46
1
$begingroup$
If the point is on the curve one can find the value of t as 1/4 of the k coefficient. Then use the formula you mentioned for a correct answer. But this point is not on the curve, so the normal to the curve cannot be found.
$endgroup$
– Peter Foreman
Jan 30 at 20:47
add a comment |
1
$begingroup$
This point is not on the curve?
$endgroup$
– Peter Foreman
Jan 30 at 20:39
$begingroup$
This point is on the curve. I couldn't find any value of $t$ for this point
$endgroup$
– Mathsaddict
Jan 30 at 20:43
$begingroup$
Therefore it is not on the curve?
$endgroup$
– Peter Foreman
Jan 30 at 20:45
$begingroup$
So, there's no solution?
$endgroup$
– Mathsaddict
Jan 30 at 20:46
1
$begingroup$
If the point is on the curve one can find the value of t as 1/4 of the k coefficient. Then use the formula you mentioned for a correct answer. But this point is not on the curve, so the normal to the curve cannot be found.
$endgroup$
– Peter Foreman
Jan 30 at 20:47
1
1
$begingroup$
This point is not on the curve?
$endgroup$
– Peter Foreman
Jan 30 at 20:39
$begingroup$
This point is not on the curve?
$endgroup$
– Peter Foreman
Jan 30 at 20:39
$begingroup$
This point is on the curve. I couldn't find any value of $t$ for this point
$endgroup$
– Mathsaddict
Jan 30 at 20:43
$begingroup$
This point is on the curve. I couldn't find any value of $t$ for this point
$endgroup$
– Mathsaddict
Jan 30 at 20:43
$begingroup$
Therefore it is not on the curve?
$endgroup$
– Peter Foreman
Jan 30 at 20:45
$begingroup$
Therefore it is not on the curve?
$endgroup$
– Peter Foreman
Jan 30 at 20:45
$begingroup$
So, there's no solution?
$endgroup$
– Mathsaddict
Jan 30 at 20:46
$begingroup$
So, there's no solution?
$endgroup$
– Mathsaddict
Jan 30 at 20:46
1
1
$begingroup$
If the point is on the curve one can find the value of t as 1/4 of the k coefficient. Then use the formula you mentioned for a correct answer. But this point is not on the curve, so the normal to the curve cannot be found.
$endgroup$
– Peter Foreman
Jan 30 at 20:47
$begingroup$
If the point is on the curve one can find the value of t as 1/4 of the k coefficient. Then use the formula you mentioned for a correct answer. But this point is not on the curve, so the normal to the curve cannot be found.
$endgroup$
– Peter Foreman
Jan 30 at 20:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Observe that your point is not on the curve, so it does not make sense to compute the unit normal at that point. Anyway, here is a general process to find the unit normal.
First of all you should parametrize your curve by arclength, which is given by
begin{align}
s(t) & = int_0^t lVert r'(x) rVert,dx \
& = 5t.
end{align}
As a function of $s$, $r$ can be rewritten as $r(s) = left(3sin frac{s}{5},3cos frac{s}{5}, frac{4}{5}sright)$. The unit tangent is then
$$T(s) = frac{1}{lVert frac{dr}{ds}rVert}frac{dr}{ds} = left(frac{3}{5}cos frac{s}{5}, -frac{3}{5}sin frac{s}{5}, frac{4}{5} right).$$
The unit normal is just given by the renormalized derivative of the tangent vector:
$$N(s)=frac{T'(s)}{lVert T'(s)rVert} = left(-sin frac{s}{5},-cos frac{s}{5}, 0right). $$
answered Jan 30 at 20:52
GibbsGibbs
5,4383927
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Observe that your point is not on the curve, so it does not make sense to compute the unit normal at that point. Anyway, here is a general process to find the unit normal.
First of all you should parametrize your curve by arclength, which is given by
begin{align}
s(t) & = int_0^t lVert r'(x) rVert,dx \
& = 5t.
end{align}
As a function of $s$, $r$ can be rewritten as $r(s) = left(3sin frac{s}{5},3cos frac{s}{5}, frac{4}{5}sright)$. The unit tangent is then
$$T(s) = frac{1}{lVert frac{dr}{ds}rVert}frac{dr}{ds} = left(frac{3}{5}cos frac{s}{5}, -frac{3}{5}sin frac{s}{5}, frac{4}{5} right).$$
The unit normal is just given by the renormalized derivative of the tangent vector:
$$N(s)=frac{T'(s)}{lVert T'(s)rVert} = left(-sin frac{s}{5},-cos frac{s}{5}, 0right). $$
answered Jan 30 at 20:52
GibbsGibbs
5,4383927
$endgroup$
add a comment |
$begingroup$
Observe that your point is not on the curve, so it does not make sense to compute the unit normal at that point. Anyway, here is a general process to find the unit normal.
First of all you should parametrize your curve by arclength, which is given by
begin{align}
s(t) & = int_0^t lVert r'(x) rVert,dx \
& = 5t.
end{align}
As a function of $s$, $r$ can be rewritten as $r(s) = left(3sin frac{s}{5},3cos frac{s}{5}, frac{4}{5}sright)$. The unit tangent is then
$$T(s) = frac{1}{lVert frac{dr}{ds}rVert}frac{dr}{ds} = left(frac{3}{5}cos frac{s}{5}, -frac{3}{5}sin frac{s}{5}, frac{4}{5} right).$$
The unit normal is just given by the renormalized derivative of the tangent vector:
$$N(s)=frac{T'(s)}{lVert T'(s)rVert} = left(-sin frac{s}{5},-cos frac{s}{5}, 0right). $$
answered Jan 30 at 20:52
GibbsGibbs
5,4383927
$endgroup$
add a comment |
$begingroup$
Observe that your point is not on the curve, so it does not make sense to compute the unit normal at that point. Anyway, here is a general process to find the unit normal.
First of all you should parametrize your curve by arclength, which is given by
begin{align}
s(t) & = int_0^t lVert r'(x) rVert,dx \
& = 5t.
end{align}
As a function of $s$, $r$ can be rewritten as $r(s) = left(3sin frac{s}{5},3cos frac{s}{5}, frac{4}{5}sright)$. The unit tangent is then
$$T(s) = frac{1}{lVert frac{dr}{ds}rVert}frac{dr}{ds} = left(frac{3}{5}cos frac{s}{5}, -frac{3}{5}sin frac{s}{5}, frac{4}{5} right).$$
The unit normal is just given by the renormalized derivative of the tangent vector:
$$N(s)=frac{T'(s)}{lVert T'(s)rVert} = left(-sin frac{s}{5},-cos frac{s}{5}, 0right). $$
answered Jan 30 at 20:52
GibbsGibbs
5,4383927
$endgroup$
Observe that your point is not on the curve, so it does not make sense to compute the unit normal at that point. Anyway, here is a general process to find the unit normal.
First of all you should parametrize your curve by arclength, which is given by
begin{align}
s(t) & = int_0^t lVert r'(x) rVert,dx \
& = 5t.
end{align}
As a function of $s$, $r$ can be rewritten as $r(s) = left(3sin frac{s}{5},3cos frac{s}{5}, frac{4}{5}sright)$. The unit tangent is then
$$T(s) = frac{1}{lVert frac{dr}{ds}rVert}frac{dr}{ds} = left(frac{3}{5}cos frac{s}{5}, -frac{3}{5}sin frac{s}{5}, frac{4}{5} right).$$
The unit normal is just given by the renormalized derivative of the tangent vector:
$$N(s)=frac{T'(s)}{lVert T'(s)rVert} = left(-sin frac{s}{5},-cos frac{s}{5}, 0right). $$
answered Jan 30 at 20:52
GibbsGibbs
5,4383927
edited Jan 30 at 20:58
answered Jan 30 at 20:52
GibbsGibbs
5,4383927
answered Jan 30 at 20:52
GibbsGibbs
5,4383927
answered Jan 30 at 20:52
GibbsGibbs
5,4383927
5,4383927
add a comment |
add a comment |
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1
$begingroup$
This point is not on the curve?
$endgroup$
– Peter Foreman
Jan 30 at 20:39
$begingroup$
This point is on the curve. I couldn't find any value of $t$ for this point
$endgroup$
– Mathsaddict
Jan 30 at 20:43
$begingroup$
Therefore it is not on the curve?
$endgroup$
– Peter Foreman
Jan 30 at 20:45
$begingroup$
So, there's no solution?
$endgroup$
– Mathsaddict
Jan 30 at 20:46
1
$begingroup$
If the point is on the curve one can find the value of t as 1/4 of the k coefficient. Then use the formula you mentioned for a correct answer. But this point is not on the curve, so the normal to the curve cannot be found.
$endgroup$
– Peter Foreman
Jan 30 at 20:47