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To find vectors if we are given a function and a point
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I have to solve the following exercise :
Find the tangent and the vertical unit vectors in the curve at the given point (four unit vectors are requested). Also design the vectors and the curves in common shape.
$1.$ $f(x)= x^2 ,$ at $(2,4)$ and
$2.$ $x^2+y^2 =6,$ at $(2,1)$
I don't know what I have to do to find the vectors, should they be in the form $vecu =aveci+bvecj$ ?I tried to find the tangent lines at the given points , will they help me to find the vectors?.I think the unit vectors should be in the form of $vecv =fracvecv$
vectors
edited Jul 20 at 12:23
Parcly Taxel
33.6k136588
asked Jul 20 at 12:21
Deppie3910
205
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up vote
1
down vote
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I have to solve the following exercise :
Find the tangent and the vertical unit vectors in the curve at the given point (four unit vectors are requested). Also design the vectors and the curves in common shape.
$1.$ $f(x)= x^2 ,$ at $(2,4)$ and
$2.$ $x^2+y^2 =6,$ at $(2,1)$
I don't know what I have to do to find the vectors, should they be in the form $vecu =aveci+bvecj$ ?I tried to find the tangent lines at the given points , will they help me to find the vectors?.I think the unit vectors should be in the form of $vecv =fracvecv$
vectors
edited Jul 20 at 12:23
Parcly Taxel
33.6k136588
asked Jul 20 at 12:21
Deppie3910
205
I don't quite get why you need to find a vertical unit vector, it's defined as vertical vector that has a magnitude of one. The tangent vector will be in the form of $vecu =aveci+bvecj$ where $veci$ is a horizontal unit vector and $vecj$ is the vertical one.
– Vasya
Jul 20 at 12:34
I don't know why,it is exactly what the excerise says.
– Deppie3910
Jul 20 at 12:48
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have to solve the following exercise :
Find the tangent and the vertical unit vectors in the curve at the given point (four unit vectors are requested). Also design the vectors and the curves in common shape.
$1.$ $f(x)= x^2 ,$ at $(2,4)$ and
$2.$ $x^2+y^2 =6,$ at $(2,1)$
I don't know what I have to do to find the vectors, should they be in the form $vecu =aveci+bvecj$ ?I tried to find the tangent lines at the given points , will they help me to find the vectors?.I think the unit vectors should be in the form of $vecv =fracvecv$
vectors
edited Jul 20 at 12:23
Parcly Taxel
33.6k136588
asked Jul 20 at 12:21
Deppie3910
205
I have to solve the following exercise :
Find the tangent and the vertical unit vectors in the curve at the given point (four unit vectors are requested). Also design the vectors and the curves in common shape.
$1.$ $f(x)= x^2 ,$ at $(2,4)$ and
$2.$ $x^2+y^2 =6,$ at $(2,1)$
I don't know what I have to do to find the vectors, should they be in the form $vecu =aveci+bvecj$ ?I tried to find the tangent lines at the given points , will they help me to find the vectors?.I think the unit vectors should be in the form of $vecv =fracvecv$
vectors
edited Jul 20 at 12:23
Parcly Taxel
33.6k136588
asked Jul 20 at 12:21
Deppie3910
205
edited Jul 20 at 12:23
Parcly Taxel
33.6k136588
edited Jul 20 at 12:23
Parcly Taxel
33.6k136588
edited Jul 20 at 12:23
Parcly Taxel
33.6k136588
33.6k136588
asked Jul 20 at 12:21
Deppie3910
205
asked Jul 20 at 12:21
Deppie3910
205
asked Jul 20 at 12:21
Deppie3910
205
205
I don't quite get why you need to find a vertical unit vector, it's defined as vertical vector that has a magnitude of one. The tangent vector will be in the form of $vecu =aveci+bvecj$ where $veci$ is a horizontal unit vector and $vecj$ is the vertical one.
– Vasya
Jul 20 at 12:34
I don't know why,it is exactly what the excerise says.
– Deppie3910
Jul 20 at 12:48
add a comment |Â
I don't quite get why you need to find a vertical unit vector, it's defined as vertical vector that has a magnitude of one. The tangent vector will be in the form of $vecu =aveci+bvecj$ where $veci$ is a horizontal unit vector and $vecj$ is the vertical one.
– Vasya
Jul 20 at 12:34
I don't know why,it is exactly what the excerise says.
– Deppie3910
Jul 20 at 12:48
I don't quite get why you need to find a vertical unit vector, it's defined as vertical vector that has a magnitude of one. The tangent vector will be in the form of $vecu =aveci+bvecj$ where $veci$ is a horizontal unit vector and $vecj$ is the vertical one.
– Vasya
Jul 20 at 12:34
I don't quite get why you need to find a vertical unit vector, it's defined as vertical vector that has a magnitude of one. The tangent vector will be in the form of $vecu =aveci+bvecj$ where $veci$ is a horizontal unit vector and $vecj$ is the vertical one.
– Vasya
Jul 20 at 12:34
I don't know why,it is exactly what the excerise says.
– Deppie3910
Jul 20 at 12:48
I don't know why,it is exactly what the excerise says.
– Deppie3910
Jul 20 at 12:48
add a comment |Â
1 Answer
1
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accepted
HINT
For the first we have
- $f'(x)=2x implies f'(2)=4 implies T=(1,4) quad N=(4,-1)$
For the second we have
- $x^2+y^2=6 implies 2xdx+2ydy=0 quad fracdydx=-frac x y implies T=(-2,1) quad N=(1,2)$
Then find the unitary vectors $T=fracT$ and $N=fracNN$.
answered Jul 20 at 12:35
gimusi
65.4k73584
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
HINT
For the first we have
- $f'(x)=2x implies f'(2)=4 implies T=(1,4) quad N=(4,-1)$
For the second we have
- $x^2+y^2=6 implies 2xdx+2ydy=0 quad fracdydx=-frac x y implies T=(-2,1) quad N=(1,2)$
Then find the unitary vectors $T=fracT$ and $N=fracNN$.
answered Jul 20 at 12:35
gimusi
65.4k73584
add a comment |Â
up vote
0
down vote
accepted
HINT
For the first we have
- $f'(x)=2x implies f'(2)=4 implies T=(1,4) quad N=(4,-1)$
For the second we have
- $x^2+y^2=6 implies 2xdx+2ydy=0 quad fracdydx=-frac x y implies T=(-2,1) quad N=(1,2)$
Then find the unitary vectors $T=fracT$ and $N=fracNN$.
answered Jul 20 at 12:35
gimusi
65.4k73584
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
HINT
For the first we have
- $f'(x)=2x implies f'(2)=4 implies T=(1,4) quad N=(4,-1)$
For the second we have
- $x^2+y^2=6 implies 2xdx+2ydy=0 quad fracdydx=-frac x y implies T=(-2,1) quad N=(1,2)$
Then find the unitary vectors $T=fracT$ and $N=fracNN$.
answered Jul 20 at 12:35
gimusi
65.4k73584
HINT
For the first we have
- $f'(x)=2x implies f'(2)=4 implies T=(1,4) quad N=(4,-1)$
For the second we have
- $x^2+y^2=6 implies 2xdx+2ydy=0 quad fracdydx=-frac x y implies T=(-2,1) quad N=(1,2)$
Then find the unitary vectors $T=fracT$ and $N=fracNN$.
answered Jul 20 at 12:35
gimusi
65.4k73584
answered Jul 20 at 12:35
gimusi
65.4k73584
answered Jul 20 at 12:35
gimusi
65.4k73584
answered Jul 20 at 12:35
gimusi
65.4k73584
65.4k73584
add a comment |Â
add a comment |Â
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I don't quite get why you need to find a vertical unit vector, it's defined as vertical vector that has a magnitude of one. The tangent vector will be in the form of $vecu =aveci+bvecj$ where $veci$ is a horizontal unit vector and $vecj$ is the vertical one.
– Vasya
Jul 20 at 12:34
I don't know why,it is exactly what the excerise says.
– Deppie3910
Jul 20 at 12:48