Mixing
Solving an equation with vector coefficients
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I want to solve $(ct)^2 = d(t)cdot d(t)$ for $t$, where $ d(t) = frac12at^2 + vt + r$
Where$ a, v$, and $r$ are all 3-dimensional vectors in Cartesian coordinates. How can I do this?
equation-solving tensors algebra
edited 10 mins ago
m_goldberg
82.5k869190
asked 3 hours ago
Sean McAllister
1134
New contributor
Sean McAllister is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
favorite
I want to solve $(ct)^2 = d(t)cdot d(t)$ for $t$, where $ d(t) = frac12at^2 + vt + r$
Where$ a, v$, and $r$ are all 3-dimensional vectors in Cartesian coordinates. How can I do this?
equation-solving tensors algebra
edited 10 mins ago
m_goldberg
82.5k869190
asked 3 hours ago
Sean McAllister
1134
New contributor
Sean McAllister is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What is the variable that you want to solve for?
– Henrik Schumacher
3 hours ago
(ct)^2 is a constant?
– Ulrich Neumann
3 hours ago
t is the variable to solve for, c is the speed of light so a constant yes.
– Sean McAllister
3 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I want to solve $(ct)^2 = d(t)cdot d(t)$ for $t$, where $ d(t) = frac12at^2 + vt + r$
Where$ a, v$, and $r$ are all 3-dimensional vectors in Cartesian coordinates. How can I do this?
equation-solving tensors algebra
edited 10 mins ago
m_goldberg
82.5k869190
asked 3 hours ago
Sean McAllister
1134
New contributor
Sean McAllister is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I want to solve $(ct)^2 = d(t)cdot d(t)$ for $t$, where $ d(t) = frac12at^2 + vt + r$
Where$ a, v$, and $r$ are all 3-dimensional vectors in Cartesian coordinates. How can I do this?
equation-solving tensors algebra
equation-solving tensors algebra
edited 10 mins ago
m_goldberg
82.5k869190
asked 3 hours ago
Sean McAllister
1134
New contributor
Sean McAllister is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 10 mins ago
m_goldberg
82.5k869190
asked 3 hours ago
Sean McAllister
1134
New contributor
Sean McAllister is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 10 mins ago
m_goldberg
82.5k869190
edited 10 mins ago
m_goldberg
82.5k869190
edited 10 mins ago
m_goldberg
82.5k869190
82.5k869190
asked 3 hours ago
Sean McAllister
1134
New contributor
Sean McAllister is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Sean McAllister
1134
asked 3 hours ago
Sean McAllister
1134
1134
New contributor
Sean McAllister is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sean McAllister is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sean McAllister is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What is the variable that you want to solve for?
– Henrik Schumacher
3 hours ago
(ct)^2 is a constant?
– Ulrich Neumann
3 hours ago
t is the variable to solve for, c is the speed of light so a constant yes.
– Sean McAllister
3 hours ago
add a comment |Â
What is the variable that you want to solve for?
– Henrik Schumacher
3 hours ago
(ct)^2 is a constant?
– Ulrich Neumann
3 hours ago
t is the variable to solve for, c is the speed of light so a constant yes.
– Sean McAllister
3 hours ago
What is the variable that you want to solve for?
– Henrik Schumacher
3 hours ago
What is the variable that you want to solve for?
– Henrik Schumacher
3 hours ago
(ct)^2 is a constant?
– Ulrich Neumann
3 hours ago
(ct)^2 is a constant?
– Ulrich Neumann
3 hours ago
t is the variable to solve for, c is the speed of light so a constant yes.
– Sean McAllister
3 hours ago
t is the variable to solve for, c is the speed of light so a constant yes.
– Sean McAllister
3 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
If you define the three vectors explicitly as
A = a1, a2, a3;
V = v1, v2, v3;
R = r1, r2, r3;
your equation evaluates to a polynom in t of order 4
eq = (c t)^2 == #.# &[1/2 A t^2 + V t + R] // Collect[#, t] &
which you might solve using MMA Solve[eq, t]
addendum
If you want preserve the invariant scalarproducts, the equation could be defined as follows
Clear[A, V, R]
(c t)^2 == 1/2 t^2, t, 1.Outer[Dot, A, V, R, A, V, R].1/2 t^2,t, 1
(*c^2 t^2 ==1/2 t^2 A.R + R.R + 1/2 t^2 (1/2 t^2 A.A + R.A + t V.A) + t V.R+t (1/2 t^2 A.V + R.V + t V.V)*)
answered 3 hours ago
Ulrich Neumann
5,080413
add a comment |Â
up vote
3
down vote
You can use TensorExpand. First, some assumptions, and your distance function:
$Assumptions = (a|v|r) ∈ Vectors[3];
d[t_] := 1/2 a t^2 + v t + r
Then, use Solve on the tensor expanded equation:
Solve[
TensorExpand[d[t] . d[t] == (c t)^2],
t,
Quartics->False
]
t -> Root[
4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 1], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 2], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 3], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 4]
Without the Quartics option, you get a mess of hard to understand radicals.
answered 3 hours ago
Carl Woll
60k279154
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
If you define the three vectors explicitly as
A = a1, a2, a3;
V = v1, v2, v3;
R = r1, r2, r3;
your equation evaluates to a polynom in t of order 4
eq = (c t)^2 == #.# &[1/2 A t^2 + V t + R] // Collect[#, t] &
which you might solve using MMA Solve[eq, t]
addendum
If you want preserve the invariant scalarproducts, the equation could be defined as follows
Clear[A, V, R]
(c t)^2 == 1/2 t^2, t, 1.Outer[Dot, A, V, R, A, V, R].1/2 t^2,t, 1
(*c^2 t^2 ==1/2 t^2 A.R + R.R + 1/2 t^2 (1/2 t^2 A.A + R.A + t V.A) + t V.R+t (1/2 t^2 A.V + R.V + t V.V)*)
answered 3 hours ago
Ulrich Neumann
5,080413
add a comment |Â
up vote
4
down vote
accepted
If you define the three vectors explicitly as
A = a1, a2, a3;
V = v1, v2, v3;
R = r1, r2, r3;
your equation evaluates to a polynom in t of order 4
eq = (c t)^2 == #.# &[1/2 A t^2 + V t + R] // Collect[#, t] &
which you might solve using MMA Solve[eq, t]
addendum
If you want preserve the invariant scalarproducts, the equation could be defined as follows
Clear[A, V, R]
(c t)^2 == 1/2 t^2, t, 1.Outer[Dot, A, V, R, A, V, R].1/2 t^2,t, 1
(*c^2 t^2 ==1/2 t^2 A.R + R.R + 1/2 t^2 (1/2 t^2 A.A + R.A + t V.A) + t V.R+t (1/2 t^2 A.V + R.V + t V.V)*)
answered 3 hours ago
Ulrich Neumann
5,080413
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
If you define the three vectors explicitly as
A = a1, a2, a3;
V = v1, v2, v3;
R = r1, r2, r3;
your equation evaluates to a polynom in t of order 4
eq = (c t)^2 == #.# &[1/2 A t^2 + V t + R] // Collect[#, t] &
which you might solve using MMA Solve[eq, t]
addendum
If you want preserve the invariant scalarproducts, the equation could be defined as follows
Clear[A, V, R]
(c t)^2 == 1/2 t^2, t, 1.Outer[Dot, A, V, R, A, V, R].1/2 t^2,t, 1
(*c^2 t^2 ==1/2 t^2 A.R + R.R + 1/2 t^2 (1/2 t^2 A.A + R.A + t V.A) + t V.R+t (1/2 t^2 A.V + R.V + t V.V)*)
answered 3 hours ago
Ulrich Neumann
5,080413
If you define the three vectors explicitly as
A = a1, a2, a3;
V = v1, v2, v3;
R = r1, r2, r3;
your equation evaluates to a polynom in t of order 4
eq = (c t)^2 == #.# &[1/2 A t^2 + V t + R] // Collect[#, t] &
which you might solve using MMA Solve[eq, t]
addendum
If you want preserve the invariant scalarproducts, the equation could be defined as follows
Clear[A, V, R]
(c t)^2 == 1/2 t^2, t, 1.Outer[Dot, A, V, R, A, V, R].1/2 t^2,t, 1
(*c^2 t^2 ==1/2 t^2 A.R + R.R + 1/2 t^2 (1/2 t^2 A.A + R.A + t V.A) + t V.R+t (1/2 t^2 A.V + R.V + t V.V)*)
answered 3 hours ago
Ulrich Neumann
5,080413
edited 2 hours ago
answered 3 hours ago
Ulrich Neumann
5,080413
answered 3 hours ago
Ulrich Neumann
5,080413
answered 3 hours ago
Ulrich Neumann
5,080413
5,080413
add a comment |Â
add a comment |Â
up vote
3
down vote
You can use TensorExpand. First, some assumptions, and your distance function:
$Assumptions = (a|v|r) ∈ Vectors[3];
d[t_] := 1/2 a t^2 + v t + r
Then, use Solve on the tensor expanded equation:
Solve[
TensorExpand[d[t] . d[t] == (c t)^2],
t,
Quartics->False
]
t -> Root[
4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 1], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 2], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 3], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 4]
Without the Quartics option, you get a mess of hard to understand radicals.
answered 3 hours ago
Carl Woll
60k279154
add a comment |Â
up vote
3
down vote
You can use TensorExpand. First, some assumptions, and your distance function:
$Assumptions = (a|v|r) ∈ Vectors[3];
d[t_] := 1/2 a t^2 + v t + r
Then, use Solve on the tensor expanded equation:
Solve[
TensorExpand[d[t] . d[t] == (c t)^2],
t,
Quartics->False
]
t -> Root[
4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 1], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 2], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 3], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 4]
Without the Quartics option, you get a mess of hard to understand radicals.
answered 3 hours ago
Carl Woll
60k279154
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You can use TensorExpand. First, some assumptions, and your distance function:
$Assumptions = (a|v|r) ∈ Vectors[3];
d[t_] := 1/2 a t^2 + v t + r
Then, use Solve on the tensor expanded equation:
Solve[
TensorExpand[d[t] . d[t] == (c t)^2],
t,
Quartics->False
]
t -> Root[
4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 1], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 2], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 3], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 4]
Without the Quartics option, you get a mess of hard to understand radicals.
answered 3 hours ago
Carl Woll
60k279154
You can use TensorExpand. First, some assumptions, and your distance function:
$Assumptions = (a|v|r) ∈ Vectors[3];
d[t_] := 1/2 a t^2 + v t + r
Then, use Solve on the tensor expanded equation:
Solve[
TensorExpand[d[t] . d[t] == (c t)^2],
t,
Quartics->False
]
t -> Root[
4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 1], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 2], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 3], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 4]
Without the Quartics option, you get a mess of hard to understand radicals.
answered 3 hours ago
Carl Woll
60k279154
answered 3 hours ago
Carl Woll
60k279154
answered 3 hours ago
Carl Woll
60k279154
answered 3 hours ago
Carl Woll
60k279154
60k279154
add a comment |Â
add a comment |Â
Sean McAllister is a new contributor. Be nice, and check out our Code of Conduct.
Sean McAllister is a new contributor. Be nice, and check out our Code of Conduct.
Sean McAllister is a new contributor. Be nice, and check out our Code of Conduct.
Sean McAllister is a new contributor. Be nice, and check out our Code of Conduct.
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f183403%2fsolving-an-equation-with-vector-coefficients%23new-answer', 'question_page');
);
Post as a guest
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
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
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
What is the variable that you want to solve for?
– Henrik Schumacher
3 hours ago
(ct)^2 is a constant?
– Ulrich Neumann
3 hours ago
t is the variable to solve for, c is the speed of light so a constant yes.
– Sean McAllister
3 hours ago