Meaning of “Derive cosine of $theta$ from $a·b$” ? (Not an native english speaker)












2












$begingroup$


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 ?










share|cite|improve this question











$endgroup$








  • 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












  • $begingroup$
    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
















2












$begingroup$


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 ?










share|cite|improve this question











$endgroup$








  • 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












  • $begingroup$
    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














2












2








2





$begingroup$


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 ?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '18 at 14:27









mrtaurho

5,53151440




5,53151440










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














  • 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










2 Answers
2






active

oldest

votes


















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















4












$begingroup$

The definition of the dot-product is given by
$$vec{a}cdot vec{b}=|vec{a}||vec{b}|cosangle(vec{a},vec{b})$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052970%2fmeaning-of-derive-cosine-of-theta-from-a-b-not-an-native-english-speak%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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











4












$begingroup$

The definition of the dot-product is given by
$$vec{a}cdot vec{b}=|vec{a}||vec{b}|cosangle(vec{a},vec{b})$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















4












$begingroup$

The definition of the dot-product is given by
$$vec{a}cdot vec{b}=|vec{a}||vec{b}|cosangle(vec{a},vec{b})$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














4












4








4





$begingroup$

The definition of the dot-product is given by
$$vec{a}cdot vec{b}=|vec{a}||vec{b}|cosangle(vec{a},vec{b})$$






share|cite|improve this answer









$endgroup$



The definition of the dot-product is given by
$$vec{a}cdot vec{b}=|vec{a}||vec{b}|cosangle(vec{a},vec{b})$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052970%2fmeaning-of-derive-cosine-of-theta-from-a-b-not-an-native-english-speak%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Ellipse (mathématiques)

Quarter-circle Tiles

Mont Emei