Meaning of “Derive cosine of $theta$ from $a·b$” ? (Not an native english speaker)
$begingroup$
I'm currently trying to implement some vehicle physics in a game, and this obviously requires a lot of maths. However, I'm not an english native speaker, so I have trouble understanding some terms and instructions that are given to me.
In my case, this is what they asked me to do :
Derive cosine of the rotation angle from the dot product of $a$ and $b$.
I assume that $theta$ here is the "rotation angle" they are talking about, and $a$ and $b$ are both vectors.
But what do they mean by "Derive... from..." ? Surely they are not asking me to calculate the derivative of $cos theta$, because they would instruct me to calculate "$-sin theta$" instead.
Can anyone please enlighten me ?
trigonometry terminology
edited Dec 26 '18 at 14:27
mrtaurho
5,53151440
asked Dec 26 '18 at 14:16
Benoît LyBenoît Ly
132
$endgroup$
add a comment |
$begingroup$
I'm currently trying to implement some vehicle physics in a game, and this obviously requires a lot of maths. However, I'm not an english native speaker, so I have trouble understanding some terms and instructions that are given to me.
In my case, this is what they asked me to do :
Derive cosine of the rotation angle from the dot product of $a$ and $b$.
I assume that $theta$ here is the "rotation angle" they are talking about, and $a$ and $b$ are both vectors.
But what do they mean by "Derive... from..." ? Surely they are not asking me to calculate the derivative of $cos theta$, because they would instruct me to calculate "$-sin theta$" instead.
Can anyone please enlighten me ?
trigonometry terminology
edited Dec 26 '18 at 14:27
mrtaurho
5,53151440
asked Dec 26 '18 at 14:16
Benoît LyBenoît Ly
132
$endgroup$
3
$begingroup$
The verb that means "to compute the derivative" is "to differentiate". "Derive" here means "obtain", "find".
$endgroup$
– ajotatxe
Dec 26 '18 at 14:18
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Hint: use the cosine rule for triangles on the triangle defined by vectors $a$ and $b$.
$endgroup$
– Mark Bennet
Dec 26 '18 at 14:25
1
$begingroup$
“Deriving” is not the same as “finding the derivative.” That’s “differentiating.” To derive means to obtain something. (It has nothing to do with derivatives.)
$endgroup$
– KM101
Dec 26 '18 at 14:26
$begingroup$
I should have known that derive and derivatives are not the same thing. Thanks for getting me out of the confusion.
$endgroup$
– Benoît Ly
Dec 26 '18 at 15:25
add a comment |
$begingroup$
I'm currently trying to implement some vehicle physics in a game, and this obviously requires a lot of maths. However, I'm not an english native speaker, so I have trouble understanding some terms and instructions that are given to me.
In my case, this is what they asked me to do :
Derive cosine of the rotation angle from the dot product of $a$ and $b$.
I assume that $theta$ here is the "rotation angle" they are talking about, and $a$ and $b$ are both vectors.
But what do they mean by "Derive... from..." ? Surely they are not asking me to calculate the derivative of $cos theta$, because they would instruct me to calculate "$-sin theta$" instead.
Can anyone please enlighten me ?
trigonometry terminology
edited Dec 26 '18 at 14:27
mrtaurho
5,53151440
asked Dec 26 '18 at 14:16
Benoît LyBenoît Ly
132
$endgroup$
I'm currently trying to implement some vehicle physics in a game, and this obviously requires a lot of maths. However, I'm not an english native speaker, so I have trouble understanding some terms and instructions that are given to me.
In my case, this is what they asked me to do :
Derive cosine of the rotation angle from the dot product of $a$ and $b$.
I assume that $theta$ here is the "rotation angle" they are talking about, and $a$ and $b$ are both vectors.
But what do they mean by "Derive... from..." ? Surely they are not asking me to calculate the derivative of $cos theta$, because they would instruct me to calculate "$-sin theta$" instead.
Can anyone please enlighten me ?
trigonometry terminology
trigonometry terminology
edited Dec 26 '18 at 14:27
mrtaurho
5,53151440
asked Dec 26 '18 at 14:16
Benoît LyBenoît Ly
132
edited Dec 26 '18 at 14:27
mrtaurho
5,53151440
asked Dec 26 '18 at 14:16
Benoît LyBenoît Ly
132
edited Dec 26 '18 at 14:27
mrtaurho
5,53151440
edited Dec 26 '18 at 14:27
mrtaurho
5,53151440
edited Dec 26 '18 at 14:27
mrtaurho
5,53151440
5,53151440
asked Dec 26 '18 at 14:16
Benoît LyBenoît Ly
132
asked Dec 26 '18 at 14:16
Benoît LyBenoît Ly
132
asked Dec 26 '18 at 14:16
Benoît LyBenoît Ly
132
132
3
$begingroup$
The verb that means "to compute the derivative" is "to differentiate". "Derive" here means "obtain", "find".
$endgroup$
– ajotatxe
Dec 26 '18 at 14:18
$begingroup$
Hint: use the cosine rule for triangles on the triangle defined by vectors $a$ and $b$.
$endgroup$
– Mark Bennet
Dec 26 '18 at 14:25
1
$begingroup$
“Deriving” is not the same as “finding the derivative.” That’s “differentiating.” To derive means to obtain something. (It has nothing to do with derivatives.)
$endgroup$
– KM101
Dec 26 '18 at 14:26
$begingroup$
I should have known that derive and derivatives are not the same thing. Thanks for getting me out of the confusion.
$endgroup$
– Benoît Ly
Dec 26 '18 at 15:25
add a comment |
3
$begingroup$
The verb that means "to compute the derivative" is "to differentiate". "Derive" here means "obtain", "find".
$endgroup$
– ajotatxe
Dec 26 '18 at 14:18
$begingroup$
Hint: use the cosine rule for triangles on the triangle defined by vectors $a$ and $b$.
$endgroup$
– Mark Bennet
Dec 26 '18 at 14:25
1
$begingroup$
“Deriving” is not the same as “finding the derivative.” That’s “differentiating.” To derive means to obtain something. (It has nothing to do with derivatives.)
$endgroup$
– KM101
Dec 26 '18 at 14:26
$begingroup$
I should have known that derive and derivatives are not the same thing. Thanks for getting me out of the confusion.
$endgroup$
– Benoît Ly
Dec 26 '18 at 15:25
3
3
$begingroup$
The verb that means "to compute the derivative" is "to differentiate". "Derive" here means "obtain", "find".
$endgroup$
– ajotatxe
Dec 26 '18 at 14:18
$begingroup$
The verb that means "to compute the derivative" is "to differentiate". "Derive" here means "obtain", "find".
$endgroup$
– ajotatxe
Dec 26 '18 at 14:18
$begingroup$
Hint: use the cosine rule for triangles on the triangle defined by vectors $a$ and $b$.
$endgroup$
– Mark Bennet
Dec 26 '18 at 14:25
$begingroup$
Hint: use the cosine rule for triangles on the triangle defined by vectors $a$ and $b$.
$endgroup$
– Mark Bennet
Dec 26 '18 at 14:25
1
1
$begingroup$
“Deriving” is not the same as “finding the derivative.” That’s “differentiating.” To derive means to obtain something. (It has nothing to do with derivatives.)
$endgroup$
– KM101
Dec 26 '18 at 14:26
$begingroup$
“Deriving” is not the same as “finding the derivative.” That’s “differentiating.” To derive means to obtain something. (It has nothing to do with derivatives.)
$endgroup$
– KM101
Dec 26 '18 at 14:26
$begingroup$
I should have known that derive and derivatives are not the same thing. Thanks for getting me out of the confusion.
$endgroup$
– Benoît Ly
Dec 26 '18 at 15:25
$begingroup$
I should have known that derive and derivatives are not the same thing. Thanks for getting me out of the confusion.
$endgroup$
– Benoît Ly
Dec 26 '18 at 15:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since there is an answer/hint already given, here is a different one, which begins in a different place and depends on the fact that the scalar product is symmetric and bilinear and is defined in such a way that $acdot a = |a|^2$ for any vector $a$.
Examine the triangle $OAB$ with $OA=a$ and $OB=b$ and the third side being $b-a$. Applying the cosine rule gives $$|b-a|^2=|a|^2+|b|^2-2|a||b|cos theta$$
Since $a.a=|a|^2$ we have $$(a-b)cdot(a-b)=acdot a+bcdot b-2|a||b|cos theta$$
and this reduces to $acdot b=|a||b|cos theta$
This allows you to use the components with respect to a suitable (orthonormal) basis to calculate everything - it demonstrates that the two different common ways of expressing the scalar product give the same result, and enables the angle to be computed using a definition which doesn't depend on knowing the angle.
answered Dec 26 '18 at 14:40
Mark BennetMark Bennet
81.4k983180
$endgroup$
$begingroup$
So after reading all these answers, this means all I have left to do is to calculate cos θ = (a⋅b) / (|a||b|). Am I right ?
$endgroup$
– Benoît Ly
Dec 26 '18 at 15:23
$begingroup$
@BenoîtLy correct
$endgroup$
– Mark Bennet
Dec 26 '18 at 16:43
$begingroup$
Thank you, your explanations are really helpful.
$endgroup$
– Benoît Ly
Dec 26 '18 at 17:16
add a comment |
$begingroup$
The definition of the dot-product is given by
$$vec{a}cdot vec{b}=|vec{a}||vec{b}|cosangle(vec{a},vec{b})$$
answered Dec 26 '18 at 14:21
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
$endgroup$
$begingroup$
If you define the scalar product in this way, you have to show and justify a way of computing it which doesn't depend on knowing the angle.
$endgroup$
– Mark Bennet
Dec 26 '18 at 14:41
$begingroup$
See here mathworld.wolfram.com/DotProduct.html
$endgroup$
– Dr. Sonnhard Graubner
Dec 26 '18 at 14:49
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since there is an answer/hint already given, here is a different one, which begins in a different place and depends on the fact that the scalar product is symmetric and bilinear and is defined in such a way that $acdot a = |a|^2$ for any vector $a$.
Examine the triangle $OAB$ with $OA=a$ and $OB=b$ and the third side being $b-a$. Applying the cosine rule gives $$|b-a|^2=|a|^2+|b|^2-2|a||b|cos theta$$
Since $a.a=|a|^2$ we have $$(a-b)cdot(a-b)=acdot a+bcdot b-2|a||b|cos theta$$
and this reduces to $acdot b=|a||b|cos theta$
This allows you to use the components with respect to a suitable (orthonormal) basis to calculate everything - it demonstrates that the two different common ways of expressing the scalar product give the same result, and enables the angle to be computed using a definition which doesn't depend on knowing the angle.
answered Dec 26 '18 at 14:40
Mark BennetMark Bennet
81.4k983180
$endgroup$
$begingroup$
So after reading all these answers, this means all I have left to do is to calculate cos θ = (a⋅b) / (|a||b|). Am I right ?
$endgroup$
– Benoît Ly
Dec 26 '18 at 15:23
$begingroup$
@BenoîtLy correct
$endgroup$
– Mark Bennet
Dec 26 '18 at 16:43
$begingroup$
Thank you, your explanations are really helpful.
$endgroup$
– Benoît Ly
Dec 26 '18 at 17:16
add a comment |
$begingroup$
Since there is an answer/hint already given, here is a different one, which begins in a different place and depends on the fact that the scalar product is symmetric and bilinear and is defined in such a way that $acdot a = |a|^2$ for any vector $a$.
Examine the triangle $OAB$ with $OA=a$ and $OB=b$ and the third side being $b-a$. Applying the cosine rule gives $$|b-a|^2=|a|^2+|b|^2-2|a||b|cos theta$$
Since $a.a=|a|^2$ we have $$(a-b)cdot(a-b)=acdot a+bcdot b-2|a||b|cos theta$$
and this reduces to $acdot b=|a||b|cos theta$
This allows you to use the components with respect to a suitable (orthonormal) basis to calculate everything - it demonstrates that the two different common ways of expressing the scalar product give the same result, and enables the angle to be computed using a definition which doesn't depend on knowing the angle.
answered Dec 26 '18 at 14:40
Mark BennetMark Bennet
81.4k983180
$endgroup$
$begingroup$
So after reading all these answers, this means all I have left to do is to calculate cos θ = (a⋅b) / (|a||b|). Am I right ?
$endgroup$
– Benoît Ly
Dec 26 '18 at 15:23
$begingroup$
@BenoîtLy correct
$endgroup$
– Mark Bennet
Dec 26 '18 at 16:43
$begingroup$
Thank you, your explanations are really helpful.
$endgroup$
– Benoît Ly
Dec 26 '18 at 17:16
add a comment |
$begingroup$
Since there is an answer/hint already given, here is a different one, which begins in a different place and depends on the fact that the scalar product is symmetric and bilinear and is defined in such a way that $acdot a = |a|^2$ for any vector $a$.
Examine the triangle $OAB$ with $OA=a$ and $OB=b$ and the third side being $b-a$. Applying the cosine rule gives $$|b-a|^2=|a|^2+|b|^2-2|a||b|cos theta$$
Since $a.a=|a|^2$ we have $$(a-b)cdot(a-b)=acdot a+bcdot b-2|a||b|cos theta$$
and this reduces to $acdot b=|a||b|cos theta$
This allows you to use the components with respect to a suitable (orthonormal) basis to calculate everything - it demonstrates that the two different common ways of expressing the scalar product give the same result, and enables the angle to be computed using a definition which doesn't depend on knowing the angle.
answered Dec 26 '18 at 14:40
Mark BennetMark Bennet
81.4k983180
$endgroup$
Since there is an answer/hint already given, here is a different one, which begins in a different place and depends on the fact that the scalar product is symmetric and bilinear and is defined in such a way that $acdot a = |a|^2$ for any vector $a$.
Examine the triangle $OAB$ with $OA=a$ and $OB=b$ and the third side being $b-a$. Applying the cosine rule gives $$|b-a|^2=|a|^2+|b|^2-2|a||b|cos theta$$
Since $a.a=|a|^2$ we have $$(a-b)cdot(a-b)=acdot a+bcdot b-2|a||b|cos theta$$
and this reduces to $acdot b=|a||b|cos theta$
This allows you to use the components with respect to a suitable (orthonormal) basis to calculate everything - it demonstrates that the two different common ways of expressing the scalar product give the same result, and enables the angle to be computed using a definition which doesn't depend on knowing the angle.
answered Dec 26 '18 at 14:40
Mark BennetMark Bennet
81.4k983180
answered Dec 26 '18 at 14:40
Mark BennetMark Bennet
81.4k983180
answered Dec 26 '18 at 14:40
Mark BennetMark Bennet
81.4k983180
answered Dec 26 '18 at 14:40
Mark BennetMark Bennet
81.4k983180
81.4k983180
$begingroup$
So after reading all these answers, this means all I have left to do is to calculate cos θ = (a⋅b) / (|a||b|). Am I right ?
$endgroup$
– Benoît Ly
Dec 26 '18 at 15:23
$begingroup$
@BenoîtLy correct
$endgroup$
– Mark Bennet
Dec 26 '18 at 16:43
$begingroup$
Thank you, your explanations are really helpful.
$endgroup$
– Benoît Ly
Dec 26 '18 at 17:16
add a comment |
$begingroup$
So after reading all these answers, this means all I have left to do is to calculate cos θ = (a⋅b) / (|a||b|). Am I right ?
$endgroup$
– Benoît Ly
Dec 26 '18 at 15:23
$begingroup$
@BenoîtLy correct
$endgroup$
– Mark Bennet
Dec 26 '18 at 16:43
$begingroup$
Thank you, your explanations are really helpful.
$endgroup$
– Benoît Ly
Dec 26 '18 at 17:16
$begingroup$
So after reading all these answers, this means all I have left to do is to calculate cos θ = (a⋅b) / (|a||b|). Am I right ?
$endgroup$
– Benoît Ly
Dec 26 '18 at 15:23
$begingroup$
So after reading all these answers, this means all I have left to do is to calculate cos θ = (a⋅b) / (|a||b|). Am I right ?
$endgroup$
– Benoît Ly
Dec 26 '18 at 15:23
$begingroup$
@BenoîtLy correct
$endgroup$
– Mark Bennet
Dec 26 '18 at 16:43
$begingroup$
@BenoîtLy correct
$endgroup$
– Mark Bennet
Dec 26 '18 at 16:43
$begingroup$
Thank you, your explanations are really helpful.
$endgroup$
– Benoît Ly
Dec 26 '18 at 17:16
$begingroup$
Thank you, your explanations are really helpful.
$endgroup$
– Benoît Ly
Dec 26 '18 at 17:16
add a comment |
$begingroup$
The definition of the dot-product is given by
$$vec{a}cdot vec{b}=|vec{a}||vec{b}|cosangle(vec{a},vec{b})$$
answered Dec 26 '18 at 14:21
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
$endgroup$
$begingroup$
If you define the scalar product in this way, you have to show and justify a way of computing it which doesn't depend on knowing the angle.
$endgroup$
– Mark Bennet
Dec 26 '18 at 14:41
$begingroup$
See here mathworld.wolfram.com/DotProduct.html
$endgroup$
– Dr. Sonnhard Graubner
Dec 26 '18 at 14:49
add a comment |
$begingroup$
The definition of the dot-product is given by
$$vec{a}cdot vec{b}=|vec{a}||vec{b}|cosangle(vec{a},vec{b})$$
answered Dec 26 '18 at 14:21
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
$endgroup$
$begingroup$
If you define the scalar product in this way, you have to show and justify a way of computing it which doesn't depend on knowing the angle.
$endgroup$
– Mark Bennet
Dec 26 '18 at 14:41
$begingroup$
See here mathworld.wolfram.com/DotProduct.html
$endgroup$
– Dr. Sonnhard Graubner
Dec 26 '18 at 14:49
add a comment |
$begingroup$
The definition of the dot-product is given by
$$vec{a}cdot vec{b}=|vec{a}||vec{b}|cosangle(vec{a},vec{b})$$
answered Dec 26 '18 at 14:21
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
$endgroup$
The definition of the dot-product is given by
$$vec{a}cdot vec{b}=|vec{a}||vec{b}|cosangle(vec{a},vec{b})$$
answered Dec 26 '18 at 14:21
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
answered Dec 26 '18 at 14:21
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
answered Dec 26 '18 at 14:21
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
answered Dec 26 '18 at 14:21
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
76.2k42866
$begingroup$
If you define the scalar product in this way, you have to show and justify a way of computing it which doesn't depend on knowing the angle.
$endgroup$
– Mark Bennet
Dec 26 '18 at 14:41
$begingroup$
See here mathworld.wolfram.com/DotProduct.html
$endgroup$
– Dr. Sonnhard Graubner
Dec 26 '18 at 14:49
add a comment |
$begingroup$
If you define the scalar product in this way, you have to show and justify a way of computing it which doesn't depend on knowing the angle.
$endgroup$
– Mark Bennet
Dec 26 '18 at 14:41
$begingroup$
See here mathworld.wolfram.com/DotProduct.html
$endgroup$
– Dr. Sonnhard Graubner
Dec 26 '18 at 14:49
$begingroup$
If you define the scalar product in this way, you have to show and justify a way of computing it which doesn't depend on knowing the angle.
$endgroup$
– Mark Bennet
Dec 26 '18 at 14:41
$begingroup$
If you define the scalar product in this way, you have to show and justify a way of computing it which doesn't depend on knowing the angle.
$endgroup$
– Mark Bennet
Dec 26 '18 at 14:41
$begingroup$
See here mathworld.wolfram.com/DotProduct.html
$endgroup$
– Dr. Sonnhard Graubner
Dec 26 '18 at 14:49
$begingroup$
See here mathworld.wolfram.com/DotProduct.html
$endgroup$
– Dr. Sonnhard Graubner
Dec 26 '18 at 14:49
add a comment |
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3
$begingroup$
The verb that means "to compute the derivative" is "to differentiate". "Derive" here means "obtain", "find".
$endgroup$
– ajotatxe
Dec 26 '18 at 14:18
$begingroup$
Hint: use the cosine rule for triangles on the triangle defined by vectors $a$ and $b$.
$endgroup$
– Mark Bennet
Dec 26 '18 at 14:25
1
$begingroup$
“Deriving” is not the same as “finding the derivative.” That’s “differentiating.” To derive means to obtain something. (It has nothing to do with derivatives.)
$endgroup$
– KM101
Dec 26 '18 at 14:26
$begingroup$
I should have known that derive and derivatives are not the same thing. Thanks for getting me out of the confusion.
$endgroup$
– Benoît Ly
Dec 26 '18 at 15:25