Tangents parallel to one another and finding the perpendicular of a vector function
up vote
0
down vote
favorite
We have
$$x(t)=
begin{pmatrix}
1+t \
t^2-t \
1-t^2 \
end{pmatrix}$$
Part 1
Are there $2$ points $x(t_1)$, $x(t_2)$, such that the function’s tangent vectors
at these points are parallel to each other? Find such points, or show that none
exist.
$$x'(t)=
begin{pmatrix}
1 \
2t-1 \
-2t\
end{pmatrix}$$
I tried to search online but how can I prove that they are parallel to each other if I can't even find the points? I got stumped at $2t^2-2t+1=0$
Basically I get a negative in a square root. And that would lead to no solution for t. My answer is wrong... or is the answer scheme wrong?
Part 2
Given the point $x(2)$, is there a second point such that the tangent
vectors are perpendicular to each other? Find such a point or show that it
doesn’t exist.
How can I find the second point? And to prove that they are perpendicular do I have to differentiate twice?
I would appreciate some guidance.
calculus functional-analysis derivatives vectors vector-analysis
edited Nov 14 at 8:12
Ng Chung Tak
13.6k31234
asked Nov 14 at 6:44
Naochi
346
|
show 3 more comments
up vote
0
down vote
favorite
We have
$$x(t)=
begin{pmatrix}
1+t \
t^2-t \
1-t^2 \
end{pmatrix}$$
Part 1
Are there $2$ points $x(t_1)$, $x(t_2)$, such that the function’s tangent vectors
at these points are parallel to each other? Find such points, or show that none
exist.
$$x'(t)=
begin{pmatrix}
1 \
2t-1 \
-2t\
end{pmatrix}$$
I tried to search online but how can I prove that they are parallel to each other if I can't even find the points? I got stumped at $2t^2-2t+1=0$
Basically I get a negative in a square root. And that would lead to no solution for t. My answer is wrong... or is the answer scheme wrong?
Part 2
Given the point $x(2)$, is there a second point such that the tangent
vectors are perpendicular to each other? Find such a point or show that it
doesn’t exist.
How can I find the second point? And to prove that they are perpendicular do I have to differentiate twice?
I would appreciate some guidance.
calculus functional-analysis derivatives vectors vector-analysis
edited Nov 14 at 8:12
Ng Chung Tak
13.6k31234
asked Nov 14 at 6:44
Naochi
346
How did you end up with that quadratic equation?
– amd
Nov 14 at 6:50
Have you tried using the fact that two vectors are parallel if $v_1=lambda v_2$ for some $lambda$.
– memerson
Nov 14 at 7:04
@amd Idk the word for it in english. I first found the magnitude(?) from the function. Then I Just simplified it.
– Naochi
Nov 14 at 7:18
@memerson I did not know you could do that! I will give it a go.
– Naochi
Nov 14 at 7:18
1
@memerson Since we’re in $mathbb R^3$ we can avoid introducing an extraneous variable by instead using $v_1times v_2=0$ as the condition for parallelism, which produces two independent equations.
– amd
Nov 14 at 19:38
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
We have
$$x(t)=
begin{pmatrix}
1+t \
t^2-t \
1-t^2 \
end{pmatrix}$$
Part 1
Are there $2$ points $x(t_1)$, $x(t_2)$, such that the function’s tangent vectors
at these points are parallel to each other? Find such points, or show that none
exist.
$$x'(t)=
begin{pmatrix}
1 \
2t-1 \
-2t\
end{pmatrix}$$
I tried to search online but how can I prove that they are parallel to each other if I can't even find the points? I got stumped at $2t^2-2t+1=0$
Basically I get a negative in a square root. And that would lead to no solution for t. My answer is wrong... or is the answer scheme wrong?
Part 2
Given the point $x(2)$, is there a second point such that the tangent
vectors are perpendicular to each other? Find such a point or show that it
doesn’t exist.
How can I find the second point? And to prove that they are perpendicular do I have to differentiate twice?
I would appreciate some guidance.
calculus functional-analysis derivatives vectors vector-analysis
edited Nov 14 at 8:12
Ng Chung Tak
13.6k31234
asked Nov 14 at 6:44
Naochi
346
We have
$$x(t)=
begin{pmatrix}
1+t \
t^2-t \
1-t^2 \
end{pmatrix}$$
Part 1
Are there $2$ points $x(t_1)$, $x(t_2)$, such that the function’s tangent vectors
at these points are parallel to each other? Find such points, or show that none
exist.
$$x'(t)=
begin{pmatrix}
1 \
2t-1 \
-2t\
end{pmatrix}$$
I tried to search online but how can I prove that they are parallel to each other if I can't even find the points? I got stumped at $2t^2-2t+1=0$
Basically I get a negative in a square root. And that would lead to no solution for t. My answer is wrong... or is the answer scheme wrong?
Part 2
Given the point $x(2)$, is there a second point such that the tangent
vectors are perpendicular to each other? Find such a point or show that it
doesn’t exist.
How can I find the second point? And to prove that they are perpendicular do I have to differentiate twice?
I would appreciate some guidance.
calculus functional-analysis derivatives vectors vector-analysis
calculus functional-analysis derivatives vectors vector-analysis
edited Nov 14 at 8:12
Ng Chung Tak
13.6k31234
asked Nov 14 at 6:44
Naochi
346
edited Nov 14 at 8:12
Ng Chung Tak
13.6k31234
asked Nov 14 at 6:44
Naochi
346
edited Nov 14 at 8:12
Ng Chung Tak
13.6k31234
edited Nov 14 at 8:12
Ng Chung Tak
13.6k31234
edited Nov 14 at 8:12
Ng Chung Tak
13.6k31234
13.6k31234
asked Nov 14 at 6:44
Naochi
346
asked Nov 14 at 6:44
Naochi
346
asked Nov 14 at 6:44
Naochi
346
346
How did you end up with that quadratic equation?
– amd
Nov 14 at 6:50
Have you tried using the fact that two vectors are parallel if $v_1=lambda v_2$ for some $lambda$.
– memerson
Nov 14 at 7:04
@amd Idk the word for it in english. I first found the magnitude(?) from the function. Then I Just simplified it.
– Naochi
Nov 14 at 7:18
@memerson I did not know you could do that! I will give it a go.
– Naochi
Nov 14 at 7:18
1
@memerson Since we’re in $mathbb R^3$ we can avoid introducing an extraneous variable by instead using $v_1times v_2=0$ as the condition for parallelism, which produces two independent equations.
– amd
Nov 14 at 19:38
|
show 3 more comments
How did you end up with that quadratic equation?
– amd
Nov 14 at 6:50
Have you tried using the fact that two vectors are parallel if $v_1=lambda v_2$ for some $lambda$.
– memerson
Nov 14 at 7:04
@amd Idk the word for it in english. I first found the magnitude(?) from the function. Then I Just simplified it.
– Naochi
Nov 14 at 7:18
@memerson I did not know you could do that! I will give it a go.
– Naochi
Nov 14 at 7:18
1
@memerson Since we’re in $mathbb R^3$ we can avoid introducing an extraneous variable by instead using $v_1times v_2=0$ as the condition for parallelism, which produces two independent equations.
– amd
Nov 14 at 19:38
How did you end up with that quadratic equation?
– amd
Nov 14 at 6:50
How did you end up with that quadratic equation?
– amd
Nov 14 at 6:50
Have you tried using the fact that two vectors are parallel if $v_1=lambda v_2$ for some $lambda$.
– memerson
Nov 14 at 7:04
Have you tried using the fact that two vectors are parallel if $v_1=lambda v_2$ for some $lambda$.
– memerson
Nov 14 at 7:04
@amd Idk the word for it in english. I first found the magnitude(?) from the function. Then I Just simplified it.
– Naochi
Nov 14 at 7:18
@amd Idk the word for it in english. I first found the magnitude(?) from the function. Then I Just simplified it.
– Naochi
Nov 14 at 7:18
@memerson I did not know you could do that! I will give it a go.
– Naochi
Nov 14 at 7:18
@memerson I did not know you could do that! I will give it a go.
– Naochi
Nov 14 at 7:18
1
1
@memerson Since we’re in $mathbb R^3$ we can avoid introducing an extraneous variable by instead using $v_1times v_2=0$ as the condition for parallelism, which produces two independent equations.
– amd
Nov 14 at 19:38
@memerson Since we’re in $mathbb R^3$ we can avoid introducing an extraneous variable by instead using $v_1times v_2=0$ as the condition for parallelism, which produces two independent equations.
– amd
Nov 14 at 19:38
|
show 3 more comments
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997890%2ftangents-parallel-to-one-another-and-finding-the-perpendicular-of-a-vector-funct%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
How did you end up with that quadratic equation?
– amd
Nov 14 at 6:50
Have you tried using the fact that two vectors are parallel if $v_1=lambda v_2$ for some $lambda$.
– memerson
Nov 14 at 7:04
@amd Idk the word for it in english. I first found the magnitude(?) from the function. Then I Just simplified it.
– Naochi
Nov 14 at 7:18
@memerson I did not know you could do that! I will give it a go.
– Naochi
Nov 14 at 7:18
1
@memerson Since we’re in $mathbb R^3$ we can avoid introducing an extraneous variable by instead using $v_1times v_2=0$ as the condition for parallelism, which produces two independent equations.
– amd
Nov 14 at 19:38