Mixing
How to solve an equation with vector coefficients?
Clash Royale CLAN TAG#URR8PPP
up vote
1
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?
linear-algebra algebraic-manipulation tensors algebra
edited 12 mins ago
Carl Woll
59.9k279154
asked 41 mins ago
Sean McAllister
1084
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
1
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?
linear-algebra algebraic-manipulation tensors algebra
edited 12 mins ago
Carl Woll
59.9k279154
asked 41 mins ago
Sean McAllister
1084
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
37 mins ago
(ct)^2 is a constant?
– Ulrich Neumann
36 mins ago
t is the variable to solve for, c is the speed of light so a constant yes.
– Sean McAllister
26 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
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?
linear-algebra algebraic-manipulation tensors algebra
edited 12 mins ago
Carl Woll
59.9k279154
asked 41 mins ago
Sean McAllister
1084
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?
linear-algebra algebraic-manipulation tensors algebra
linear-algebra algebraic-manipulation tensors algebra
edited 12 mins ago
Carl Woll
59.9k279154
asked 41 mins ago
Sean McAllister
1084
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 12 mins ago
Carl Woll
59.9k279154
asked 41 mins ago
Sean McAllister
1084
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 12 mins ago
Carl Woll
59.9k279154
edited 12 mins ago
Carl Woll
59.9k279154
edited 12 mins ago
Carl Woll
59.9k279154
59.9k279154
asked 41 mins ago
Sean McAllister
1084
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 41 mins ago
Sean McAllister
1084
asked 41 mins ago
Sean McAllister
1084
1084
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
37 mins ago
(ct)^2 is a constant?
– Ulrich Neumann
36 mins ago
t is the variable to solve for, c is the speed of light so a constant yes.
– Sean McAllister
26 mins ago
add a comment |Â
What is the variable that you want to solve for?
– Henrik Schumacher
37 mins ago
(ct)^2 is a constant?
– Ulrich Neumann
36 mins ago
t is the variable to solve for, c is the speed of light so a constant yes.
– Sean McAllister
26 mins ago
What is the variable that you want to solve for?
– Henrik Schumacher
37 mins ago
What is the variable that you want to solve for?
– Henrik Schumacher
37 mins ago
(ct)^2 is a constant?
– Ulrich Neumann
36 mins ago
(ct)^2 is a constant?
– Ulrich Neumann
36 mins ago
t is the variable to solve for, c is the speed of light so a constant yes.
– Sean McAllister
26 mins ago
t is the variable to solve for, c is the speed of light so a constant yes.
– Sean McAllister
26 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
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]
answered 17 mins ago
Ulrich Neumann
5,070413
add a comment |Â
up vote
2
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 13 mins ago
Carl Woll
59.9k279154
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
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]
answered 17 mins ago
Ulrich Neumann
5,070413
add a comment |Â
up vote
3
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]
answered 17 mins ago
Ulrich Neumann
5,070413
add a comment |Â
up vote
3
down vote
accepted
up vote
3
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]
answered 17 mins ago
Ulrich Neumann
5,070413
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]
answered 17 mins ago
Ulrich Neumann
5,070413
edited 4 mins ago
answered 17 mins ago
Ulrich Neumann
5,070413
answered 17 mins ago
Ulrich Neumann
5,070413
answered 17 mins ago
Ulrich Neumann
5,070413
5,070413
add a comment |Â
add a comment |Â
up vote
2
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 13 mins ago
Carl Woll
59.9k279154
add a comment |Â
up vote
2
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 13 mins ago
Carl Woll
59.9k279154
add a comment |Â
up vote
2
down vote
up vote
2
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 13 mins ago
Carl Woll
59.9k279154
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 13 mins ago
Carl Woll
59.9k279154
answered 13 mins ago
Carl Woll
59.9k279154
answered 13 mins ago
Carl Woll
59.9k279154
answered 13 mins ago
Carl Woll
59.9k279154
59.9k279154
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.
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What is the variable that you want to solve for?
– Henrik Schumacher
37 mins ago
(ct)^2 is a constant?
– Ulrich Neumann
36 mins ago
t is the variable to solve for, c is the speed of light so a constant yes.
– Sean McAllister
26 mins ago