Question about “Approaching Zero and Limits” in the Intuitive Proof of the Derivative of SineQuestion about limits involving exponents and basesSome question about sine functionsFlash question about expanding the derivative definitionThe 'sine and cosine theorem' - formulas for the sum and differenceQuestion about the derivative definitionProof of the derivative of sineQuestion about two-sided limitsEpsilon-Delta and Limits Approaching Infinity
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Question about “Approaching Zero and Limits” in the Intuitive Proof of the Derivative of Sine
Question about limits involving exponents and basesSome question about sine functionsFlash question about expanding the derivative definitionThe 'sine and cosine theorem' - formulas for the sum and differenceQuestion about the derivative definitionProof of the derivative of sineQuestion about two-sided limitsEpsilon-Delta and Limits Approaching Infinity
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$begingroup$
I'm a high school student hoping to self-study some introductory calculus over the summer. While studying, I came across this intuitive proof of the derivative of the sine function, using trig and the unit circle...
As in the picture, as dθ approaches zero, angles A and B will approach 90° -- allowing triangle ABC to be "approaching" similar to triangle BDE, but this would mean you would never get the exact angles for triangle ABC; thus, never the exact ratio of sine and cosine in order to complete the proof.
Is this small (even negligible) inaccuracy inherent to calculus, or is there a flaw in my understanding?
P.S. Please forgive me if this is a stupid/far-too-basic question.
calculus algebra-precalculus definition
edited 8 hours ago
Blue
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asked 8 hours ago
FrankFrank
232 bronze badges
New contributor
Frank is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
I'm a high school student hoping to self-study some introductory calculus over the summer. While studying, I came across this intuitive proof of the derivative of the sine function, using trig and the unit circle...
As in the picture, as dθ approaches zero, angles A and B will approach 90° -- allowing triangle ABC to be "approaching" similar to triangle BDE, but this would mean you would never get the exact angles for triangle ABC; thus, never the exact ratio of sine and cosine in order to complete the proof.
Is this small (even negligible) inaccuracy inherent to calculus, or is there a flaw in my understanding?
P.S. Please forgive me if this is a stupid/far-too-basic question.
calculus algebra-precalculus definition
edited 8 hours ago
Blue
52.2k11 gold badges74 silver badges166 bronze badges
asked 8 hours ago
FrankFrank
232 bronze badges
New contributor
Frank is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
It has to do with the definition of limit. Do you know the formal definition?
$endgroup$
– quasi
8 hours ago
1
$begingroup$
Minor comment, you forgot to label the ABCDE in the diagrams (or at least, I cannot see them since the apparently hand drawn picture is quite small).
$endgroup$
– pre-kidney
8 hours ago
add a comment |
$begingroup$
I'm a high school student hoping to self-study some introductory calculus over the summer. While studying, I came across this intuitive proof of the derivative of the sine function, using trig and the unit circle...
As in the picture, as dθ approaches zero, angles A and B will approach 90° -- allowing triangle ABC to be "approaching" similar to triangle BDE, but this would mean you would never get the exact angles for triangle ABC; thus, never the exact ratio of sine and cosine in order to complete the proof.
Is this small (even negligible) inaccuracy inherent to calculus, or is there a flaw in my understanding?
P.S. Please forgive me if this is a stupid/far-too-basic question.
calculus algebra-precalculus definition
edited 8 hours ago
Blue
52.2k11 gold badges74 silver badges166 bronze badges
asked 8 hours ago
FrankFrank
232 bronze badges
New contributor
Frank is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm a high school student hoping to self-study some introductory calculus over the summer. While studying, I came across this intuitive proof of the derivative of the sine function, using trig and the unit circle...
As in the picture, as dθ approaches zero, angles A and B will approach 90° -- allowing triangle ABC to be "approaching" similar to triangle BDE, but this would mean you would never get the exact angles for triangle ABC; thus, never the exact ratio of sine and cosine in order to complete the proof.
Is this small (even negligible) inaccuracy inherent to calculus, or is there a flaw in my understanding?
P.S. Please forgive me if this is a stupid/far-too-basic question.
calculus algebra-precalculus definition
calculus algebra-precalculus definition
edited 8 hours ago
Blue
52.2k11 gold badges74 silver badges166 bronze badges
asked 8 hours ago
FrankFrank
232 bronze badges
New contributor
Frank is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
Blue
52.2k11 gold badges74 silver badges166 bronze badges
asked 8 hours ago
FrankFrank
232 bronze badges
New contributor
Frank is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
Blue
52.2k11 gold badges74 silver badges166 bronze badges
edited 8 hours ago
Blue
52.2k11 gold badges74 silver badges166 bronze badges
edited 8 hours ago
Blue
52.2k11 gold badges74 silver badges166 bronze badges
52.2k11 gold badges74 silver badges166 bronze badges
asked 8 hours ago
FrankFrank
232 bronze badges
New contributor
Frank is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
FrankFrank
232 bronze badges
asked 8 hours ago
FrankFrank
232 bronze badges
232 bronze badges
New contributor
Frank is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Frank is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
It has to do with the definition of limit. Do you know the formal definition?
$endgroup$
– quasi
8 hours ago
1
$begingroup$
Minor comment, you forgot to label the ABCDE in the diagrams (or at least, I cannot see them since the apparently hand drawn picture is quite small).
$endgroup$
– pre-kidney
8 hours ago
add a comment |
$begingroup$
It has to do with the definition of limit. Do you know the formal definition?
$endgroup$
– quasi
8 hours ago
1
$begingroup$
Minor comment, you forgot to label the ABCDE in the diagrams (or at least, I cannot see them since the apparently hand drawn picture is quite small).
$endgroup$
– pre-kidney
8 hours ago
$begingroup$
It has to do with the definition of limit. Do you know the formal definition?
$endgroup$
– quasi
8 hours ago
$begingroup$
It has to do with the definition of limit. Do you know the formal definition?
$endgroup$
– quasi
8 hours ago
1
1
$begingroup$
Minor comment, you forgot to label the ABCDE in the diagrams (or at least, I cannot see them since the apparently hand drawn picture is quite small).
$endgroup$
– pre-kidney
8 hours ago
$begingroup$
Minor comment, you forgot to label the ABCDE in the diagrams (or at least, I cannot see them since the apparently hand drawn picture is quite small).
$endgroup$
– pre-kidney
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The inaccuracy is of lower order than the main quantities of interest, and becomes fully accurate in the limit. The key to this question is understanding precisely what the limit means, and for this I direct you towards the rigorous meaning of a limit. Note that an initial course in calculus will include many such "hand-wavy" arguments that seem to be slightly inaccurate, and it is typically in a later course (often called "analysis" or "real analysis") following calculus where this rigorous definition is presented and results in calculus are put on a more solid footing.
For a more concrete instance of this, consider an expression like $x+x^2$ in the limit as $x$ tends to $0$. Now of course, $x+x^2$ does not equal $x$, but now if you imagine substituting a super tiny value for $x$, you will see that $x$ becomes a great approximation for $x(1+x)=x+x^2$. Note that this is stronger than saying that $x$ and $x+x^2$ have the same limit as $x$ goes to $0$, since $x^2$ also has this property - yet $x^2$ is a terrible approximation for $x+x^2$ in the limit when $x$ goes to $0$, since $x^2$ is much, much, much smaller than $x$. (I know this example is somewhat contrived, but it actually includes the main ideas in a particularly simple setting.)
answered 8 hours ago
pre-kidneypre-kidney
17.4k20 silver badges56 bronze badges
$endgroup$
add a comment |
$begingroup$
Proving a calculus theorem using pictures is not a very good idea.
While you may get a vague outline from the graphs, you also get confused when concepts like limits are involved.
If you study the definition of limits and read the proof of this theorem based on the exact definition, you will get a clear understanding as to why derivative of $sin x$ is $cos x$ and why the picture does not do a good job.
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
51.1k4 gold badges27 silver badges72 bronze badges
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
The inaccuracy is of lower order than the main quantities of interest, and becomes fully accurate in the limit. The key to this question is understanding precisely what the limit means, and for this I direct you towards the rigorous meaning of a limit. Note that an initial course in calculus will include many such "hand-wavy" arguments that seem to be slightly inaccurate, and it is typically in a later course (often called "analysis" or "real analysis") following calculus where this rigorous definition is presented and results in calculus are put on a more solid footing.
For a more concrete instance of this, consider an expression like $x+x^2$ in the limit as $x$ tends to $0$. Now of course, $x+x^2$ does not equal $x$, but now if you imagine substituting a super tiny value for $x$, you will see that $x$ becomes a great approximation for $x(1+x)=x+x^2$. Note that this is stronger than saying that $x$ and $x+x^2$ have the same limit as $x$ goes to $0$, since $x^2$ also has this property - yet $x^2$ is a terrible approximation for $x+x^2$ in the limit when $x$ goes to $0$, since $x^2$ is much, much, much smaller than $x$. (I know this example is somewhat contrived, but it actually includes the main ideas in a particularly simple setting.)
answered 8 hours ago
pre-kidneypre-kidney
17.4k20 silver badges56 bronze badges
$endgroup$
add a comment |
$begingroup$
The inaccuracy is of lower order than the main quantities of interest, and becomes fully accurate in the limit. The key to this question is understanding precisely what the limit means, and for this I direct you towards the rigorous meaning of a limit. Note that an initial course in calculus will include many such "hand-wavy" arguments that seem to be slightly inaccurate, and it is typically in a later course (often called "analysis" or "real analysis") following calculus where this rigorous definition is presented and results in calculus are put on a more solid footing.
For a more concrete instance of this, consider an expression like $x+x^2$ in the limit as $x$ tends to $0$. Now of course, $x+x^2$ does not equal $x$, but now if you imagine substituting a super tiny value for $x$, you will see that $x$ becomes a great approximation for $x(1+x)=x+x^2$. Note that this is stronger than saying that $x$ and $x+x^2$ have the same limit as $x$ goes to $0$, since $x^2$ also has this property - yet $x^2$ is a terrible approximation for $x+x^2$ in the limit when $x$ goes to $0$, since $x^2$ is much, much, much smaller than $x$. (I know this example is somewhat contrived, but it actually includes the main ideas in a particularly simple setting.)
answered 8 hours ago
pre-kidneypre-kidney
17.4k20 silver badges56 bronze badges
$endgroup$
add a comment |
$begingroup$
The inaccuracy is of lower order than the main quantities of interest, and becomes fully accurate in the limit. The key to this question is understanding precisely what the limit means, and for this I direct you towards the rigorous meaning of a limit. Note that an initial course in calculus will include many such "hand-wavy" arguments that seem to be slightly inaccurate, and it is typically in a later course (often called "analysis" or "real analysis") following calculus where this rigorous definition is presented and results in calculus are put on a more solid footing.
For a more concrete instance of this, consider an expression like $x+x^2$ in the limit as $x$ tends to $0$. Now of course, $x+x^2$ does not equal $x$, but now if you imagine substituting a super tiny value for $x$, you will see that $x$ becomes a great approximation for $x(1+x)=x+x^2$. Note that this is stronger than saying that $x$ and $x+x^2$ have the same limit as $x$ goes to $0$, since $x^2$ also has this property - yet $x^2$ is a terrible approximation for $x+x^2$ in the limit when $x$ goes to $0$, since $x^2$ is much, much, much smaller than $x$. (I know this example is somewhat contrived, but it actually includes the main ideas in a particularly simple setting.)
answered 8 hours ago
pre-kidneypre-kidney
17.4k20 silver badges56 bronze badges
$endgroup$
The inaccuracy is of lower order than the main quantities of interest, and becomes fully accurate in the limit. The key to this question is understanding precisely what the limit means, and for this I direct you towards the rigorous meaning of a limit. Note that an initial course in calculus will include many such "hand-wavy" arguments that seem to be slightly inaccurate, and it is typically in a later course (often called "analysis" or "real analysis") following calculus where this rigorous definition is presented and results in calculus are put on a more solid footing.
For a more concrete instance of this, consider an expression like $x+x^2$ in the limit as $x$ tends to $0$. Now of course, $x+x^2$ does not equal $x$, but now if you imagine substituting a super tiny value for $x$, you will see that $x$ becomes a great approximation for $x(1+x)=x+x^2$. Note that this is stronger than saying that $x$ and $x+x^2$ have the same limit as $x$ goes to $0$, since $x^2$ also has this property - yet $x^2$ is a terrible approximation for $x+x^2$ in the limit when $x$ goes to $0$, since $x^2$ is much, much, much smaller than $x$. (I know this example is somewhat contrived, but it actually includes the main ideas in a particularly simple setting.)
answered 8 hours ago
pre-kidneypre-kidney
17.4k20 silver badges56 bronze badges
edited 8 hours ago
answered 8 hours ago
pre-kidneypre-kidney
17.4k20 silver badges56 bronze badges
answered 8 hours ago
pre-kidneypre-kidney
17.4k20 silver badges56 bronze badges
answered 8 hours ago
pre-kidneypre-kidney
17.4k20 silver badges56 bronze badges
17.4k20 silver badges56 bronze badges
add a comment |
add a comment |
$begingroup$
Proving a calculus theorem using pictures is not a very good idea.
While you may get a vague outline from the graphs, you also get confused when concepts like limits are involved.
If you study the definition of limits and read the proof of this theorem based on the exact definition, you will get a clear understanding as to why derivative of $sin x$ is $cos x$ and why the picture does not do a good job.
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
51.1k4 gold badges27 silver badges72 bronze badges
$endgroup$
add a comment |
$begingroup$
Proving a calculus theorem using pictures is not a very good idea.
While you may get a vague outline from the graphs, you also get confused when concepts like limits are involved.
If you study the definition of limits and read the proof of this theorem based on the exact definition, you will get a clear understanding as to why derivative of $sin x$ is $cos x$ and why the picture does not do a good job.
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
51.1k4 gold badges27 silver badges72 bronze badges
$endgroup$
add a comment |
$begingroup$
Proving a calculus theorem using pictures is not a very good idea.
While you may get a vague outline from the graphs, you also get confused when concepts like limits are involved.
If you study the definition of limits and read the proof of this theorem based on the exact definition, you will get a clear understanding as to why derivative of $sin x$ is $cos x$ and why the picture does not do a good job.
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
51.1k4 gold badges27 silver badges72 bronze badges
$endgroup$
Proving a calculus theorem using pictures is not a very good idea.
While you may get a vague outline from the graphs, you also get confused when concepts like limits are involved.
If you study the definition of limits and read the proof of this theorem based on the exact definition, you will get a clear understanding as to why derivative of $sin x$ is $cos x$ and why the picture does not do a good job.
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
51.1k4 gold badges27 silver badges72 bronze badges
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
51.1k4 gold badges27 silver badges72 bronze badges
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
51.1k4 gold badges27 silver badges72 bronze badges
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
51.1k4 gold badges27 silver badges72 bronze badges
51.1k4 gold badges27 silver badges72 bronze badges
add a comment |
add a comment |
Frank is a new contributor. Be nice, and check out our Code of Conduct.
Frank is a new contributor. Be nice, and check out our Code of Conduct.
Frank is a new contributor. Be nice, and check out our Code of Conduct.
Frank is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
It has to do with the definition of limit. Do you know the formal definition?
$endgroup$
– quasi
8 hours ago
1
$begingroup$
Minor comment, you forgot to label the ABCDE in the diagrams (or at least, I cannot see them since the apparently hand drawn picture is quite small).
$endgroup$
– pre-kidney
8 hours ago