Mixing
Geometric Proof that $frac{d}{dtheta}(tan{theta}) = 1 + tan^{2}{theta}$
$begingroup$
The following proof sketch is from the preface of Tristan Needham's Visual Complex Analysis:
Why is the left side of the black triangle labeled $L , dtheta$?
I can see that the length of this side is $L tan{dtheta}$, so it seems that Needham is approximating $tan{dtheta}$ with $dtheta$ as $dtheta$ approaches $0$. Why is this justified? It makes sense to me in light of the fact that the derivative of $tan{theta}$ at $theta = 0$ is $1$, but this doesn't seem like the intended justification, given what he's trying to prove. Is there another, more obvious (and perhaps geometric), justification?
calculus geometry
edited Jan 16 at 17:24
nicoguaro
1237
asked Jan 16 at 17:18
NoctisNoctis
84
$endgroup$
add a comment |
$begingroup$
The following proof sketch is from the preface of Tristan Needham's Visual Complex Analysis:
Why is the left side of the black triangle labeled $L , dtheta$?
I can see that the length of this side is $L tan{dtheta}$, so it seems that Needham is approximating $tan{dtheta}$ with $dtheta$ as $dtheta$ approaches $0$. Why is this justified? It makes sense to me in light of the fact that the derivative of $tan{theta}$ at $theta = 0$ is $1$, but this doesn't seem like the intended justification, given what he's trying to prove. Is there another, more obvious (and perhaps geometric), justification?
calculus geometry
edited Jan 16 at 17:24
nicoguaro
1237
asked Jan 16 at 17:18
NoctisNoctis
84
$endgroup$
add a comment |
$begingroup$
The following proof sketch is from the preface of Tristan Needham's Visual Complex Analysis:
Why is the left side of the black triangle labeled $L , dtheta$?
I can see that the length of this side is $L tan{dtheta}$, so it seems that Needham is approximating $tan{dtheta}$ with $dtheta$ as $dtheta$ approaches $0$. Why is this justified? It makes sense to me in light of the fact that the derivative of $tan{theta}$ at $theta = 0$ is $1$, but this doesn't seem like the intended justification, given what he's trying to prove. Is there another, more obvious (and perhaps geometric), justification?
calculus geometry
edited Jan 16 at 17:24
nicoguaro
1237
asked Jan 16 at 17:18
NoctisNoctis
84
$endgroup$
The following proof sketch is from the preface of Tristan Needham's Visual Complex Analysis:
Why is the left side of the black triangle labeled $L , dtheta$?
I can see that the length of this side is $L tan{dtheta}$, so it seems that Needham is approximating $tan{dtheta}$ with $dtheta$ as $dtheta$ approaches $0$. Why is this justified? It makes sense to me in light of the fact that the derivative of $tan{theta}$ at $theta = 0$ is $1$, but this doesn't seem like the intended justification, given what he's trying to prove. Is there another, more obvious (and perhaps geometric), justification?
calculus geometry
calculus geometry
edited Jan 16 at 17:24
nicoguaro
1237
asked Jan 16 at 17:18
NoctisNoctis
84
edited Jan 16 at 17:24
nicoguaro
1237
asked Jan 16 at 17:18
NoctisNoctis
84
edited Jan 16 at 17:24
nicoguaro
1237
edited Jan 16 at 17:24
nicoguaro
1237
edited Jan 16 at 17:24
nicoguaro
1237
1237
asked Jan 16 at 17:18
NoctisNoctis
84
asked Jan 16 at 17:18
NoctisNoctis
84
asked Jan 16 at 17:18
NoctisNoctis
84
84
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$Ldtheta$ is an arc length. You can see that the dotted line rotated $dtheta$ degrees.
So, the arc length, by definition, is the radius length multiplied by the degree. In this case, $dtheta$ is too small, so we can approximate it of a straight line.
answered Jan 16 at 17:25
user4642user4642
362
$endgroup$
add a comment |
$begingroup$
It was already known to ancient greek masters that $sin theta < theta < tan theta$ (polygons inscribed and circumscribed to a circle), so I suppose that Newton,and the author here, are giving that for acquainted.
answered Jan 16 at 17:35
G CabG Cab
19.6k31239
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$Ldtheta$ is an arc length. You can see that the dotted line rotated $dtheta$ degrees.
So, the arc length, by definition, is the radius length multiplied by the degree. In this case, $dtheta$ is too small, so we can approximate it of a straight line.
answered Jan 16 at 17:25
user4642user4642
362
$endgroup$
add a comment |
$begingroup$
$Ldtheta$ is an arc length. You can see that the dotted line rotated $dtheta$ degrees.
So, the arc length, by definition, is the radius length multiplied by the degree. In this case, $dtheta$ is too small, so we can approximate it of a straight line.
answered Jan 16 at 17:25
user4642user4642
362
$endgroup$
add a comment |
$begingroup$
$Ldtheta$ is an arc length. You can see that the dotted line rotated $dtheta$ degrees.
So, the arc length, by definition, is the radius length multiplied by the degree. In this case, $dtheta$ is too small, so we can approximate it of a straight line.
answered Jan 16 at 17:25
user4642user4642
362
$endgroup$
$Ldtheta$ is an arc length. You can see that the dotted line rotated $dtheta$ degrees.
So, the arc length, by definition, is the radius length multiplied by the degree. In this case, $dtheta$ is too small, so we can approximate it of a straight line.
answered Jan 16 at 17:25
user4642user4642
362
answered Jan 16 at 17:25
user4642user4642
362
answered Jan 16 at 17:25
user4642user4642
362
answered Jan 16 at 17:25
user4642user4642
362
362
add a comment |
add a comment |
$begingroup$
It was already known to ancient greek masters that $sin theta < theta < tan theta$ (polygons inscribed and circumscribed to a circle), so I suppose that Newton,and the author here, are giving that for acquainted.
answered Jan 16 at 17:35
G CabG Cab
19.6k31239
$endgroup$
add a comment |
$begingroup$
It was already known to ancient greek masters that $sin theta < theta < tan theta$ (polygons inscribed and circumscribed to a circle), so I suppose that Newton,and the author here, are giving that for acquainted.
answered Jan 16 at 17:35
G CabG Cab
19.6k31239
$endgroup$
add a comment |
$begingroup$
It was already known to ancient greek masters that $sin theta < theta < tan theta$ (polygons inscribed and circumscribed to a circle), so I suppose that Newton,and the author here, are giving that for acquainted.
answered Jan 16 at 17:35
G CabG Cab
19.6k31239
$endgroup$
It was already known to ancient greek masters that $sin theta < theta < tan theta$ (polygons inscribed and circumscribed to a circle), so I suppose that Newton,and the author here, are giving that for acquainted.
answered Jan 16 at 17:35
G CabG Cab
19.6k31239
answered Jan 16 at 17:35
G CabG Cab
19.6k31239
answered Jan 16 at 17:35
G CabG Cab
19.6k31239
answered Jan 16 at 17:35
G CabG Cab
19.6k31239
19.6k31239
add a comment |
add a comment |
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