Why doesn't a 20 degree rotation change the slopes of $y=x$ and $y=frac{x}{2}$ by the same amount?
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It seems that if I rotate different lines (lying in the same quadrant) the same number of degrees they move different amounts (in terms of their slope). (where the rotation is such that all the lines do not enter a different quadrant)
Can someone give me intuition why this is?
My guess was that maybe it has something to do with the nature of a circle.
graphing-functions intuition rotations
asked Dec 12 '18 at 1:57
user106860user106860
373315
$endgroup$
|
show 1 more comment
$begingroup$
It seems that if I rotate different lines (lying in the same quadrant) the same number of degrees they move different amounts (in terms of their slope). (where the rotation is such that all the lines do not enter a different quadrant)
Can someone give me intuition why this is?
My guess was that maybe it has something to do with the nature of a circle.
graphing-functions intuition rotations
asked Dec 12 '18 at 1:57
user106860user106860
373315
$endgroup$
7
$begingroup$
Because the slope the tangent of the angle, and the tangent function is nonlinear.
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– Matt Samuel
Dec 12 '18 at 2:00
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@MattSamuel That helps, thank you. but now I'm wondering why the tangent function is nonlinear.
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– user106860
Dec 12 '18 at 2:05
2
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Perhaps you should wonder why it would be linear!
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– The Chaz 2.0
Dec 12 '18 at 2:06
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@user Tongue in cheek, that seems more like a philosophy question to me. From my perspective, it is what it is. But perhaps someone has a more insightful answer than that.
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– Matt Samuel
Dec 12 '18 at 2:07
2
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A rotation can't add the same constant to every slope, because then it would never make a line vertical.
$endgroup$
– Robert Israel
Dec 12 '18 at 2:18
|
show 1 more comment
$begingroup$
It seems that if I rotate different lines (lying in the same quadrant) the same number of degrees they move different amounts (in terms of their slope). (where the rotation is such that all the lines do not enter a different quadrant)
Can someone give me intuition why this is?
My guess was that maybe it has something to do with the nature of a circle.
graphing-functions intuition rotations
asked Dec 12 '18 at 1:57
user106860user106860
373315
$endgroup$
It seems that if I rotate different lines (lying in the same quadrant) the same number of degrees they move different amounts (in terms of their slope). (where the rotation is such that all the lines do not enter a different quadrant)
Can someone give me intuition why this is?
My guess was that maybe it has something to do with the nature of a circle.
graphing-functions intuition rotations
graphing-functions intuition rotations
asked Dec 12 '18 at 1:57
user106860user106860
373315
asked Dec 12 '18 at 1:57
user106860user106860
373315
asked Dec 12 '18 at 1:57
user106860user106860
373315
asked Dec 12 '18 at 1:57
user106860user106860
373315
asked Dec 12 '18 at 1:57
user106860user106860
373315
373315
7
$begingroup$
Because the slope the tangent of the angle, and the tangent function is nonlinear.
$endgroup$
– Matt Samuel
Dec 12 '18 at 2:00
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@MattSamuel That helps, thank you. but now I'm wondering why the tangent function is nonlinear.
$endgroup$
– user106860
Dec 12 '18 at 2:05
2
$begingroup$
Perhaps you should wonder why it would be linear!
$endgroup$
– The Chaz 2.0
Dec 12 '18 at 2:06
$begingroup$
@user Tongue in cheek, that seems more like a philosophy question to me. From my perspective, it is what it is. But perhaps someone has a more insightful answer than that.
$endgroup$
– Matt Samuel
Dec 12 '18 at 2:07
2
$begingroup$
A rotation can't add the same constant to every slope, because then it would never make a line vertical.
$endgroup$
– Robert Israel
Dec 12 '18 at 2:18
|
show 1 more comment
7
$begingroup$
Because the slope the tangent of the angle, and the tangent function is nonlinear.
$endgroup$
– Matt Samuel
Dec 12 '18 at 2:00
$begingroup$
@MattSamuel That helps, thank you. but now I'm wondering why the tangent function is nonlinear.
$endgroup$
– user106860
Dec 12 '18 at 2:05
2
$begingroup$
Perhaps you should wonder why it would be linear!
$endgroup$
– The Chaz 2.0
Dec 12 '18 at 2:06
$begingroup$
@user Tongue in cheek, that seems more like a philosophy question to me. From my perspective, it is what it is. But perhaps someone has a more insightful answer than that.
$endgroup$
– Matt Samuel
Dec 12 '18 at 2:07
2
$begingroup$
A rotation can't add the same constant to every slope, because then it would never make a line vertical.
$endgroup$
– Robert Israel
Dec 12 '18 at 2:18
7
7
$begingroup$
Because the slope the tangent of the angle, and the tangent function is nonlinear.
$endgroup$
– Matt Samuel
Dec 12 '18 at 2:00
$begingroup$
Because the slope the tangent of the angle, and the tangent function is nonlinear.
$endgroup$
– Matt Samuel
Dec 12 '18 at 2:00
$begingroup$
@MattSamuel That helps, thank you. but now I'm wondering why the tangent function is nonlinear.
$endgroup$
– user106860
Dec 12 '18 at 2:05
$begingroup$
@MattSamuel That helps, thank you. but now I'm wondering why the tangent function is nonlinear.
$endgroup$
– user106860
Dec 12 '18 at 2:05
2
2
$begingroup$
Perhaps you should wonder why it would be linear!
$endgroup$
– The Chaz 2.0
Dec 12 '18 at 2:06
$begingroup$
Perhaps you should wonder why it would be linear!
$endgroup$
– The Chaz 2.0
Dec 12 '18 at 2:06
$begingroup$
@user Tongue in cheek, that seems more like a philosophy question to me. From my perspective, it is what it is. But perhaps someone has a more insightful answer than that.
$endgroup$
– Matt Samuel
Dec 12 '18 at 2:07
$begingroup$
@user Tongue in cheek, that seems more like a philosophy question to me. From my perspective, it is what it is. But perhaps someone has a more insightful answer than that.
$endgroup$
– Matt Samuel
Dec 12 '18 at 2:07
2
2
$begingroup$
A rotation can't add the same constant to every slope, because then it would never make a line vertical.
$endgroup$
– Robert Israel
Dec 12 '18 at 2:18
$begingroup$
A rotation can't add the same constant to every slope, because then it would never make a line vertical.
$endgroup$
– Robert Israel
Dec 12 '18 at 2:18
|
show 1 more comment
2 Answers
2
active
oldest
votes
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Slope is $tan x$ , $x$ being the angle the line makes with $x$ axis. $x$ is 45 in the line $y=x$ and around 26 when $2y=x$. As you know that $tan$ approaches infinity when $x$ approaches 90, so as we add 20 to 45, the value of $tan$ will increase at a greater rate than when we add 20 to ~26.
answered Dec 12 '18 at 2:14
RaghavRaghav
647
$endgroup$
$begingroup$
No, $tan$ approaches $infty$ at $fracpi2$ is not sufficient to prove the increase is greater. You need to prove $tan''(x)>0$.
$endgroup$
– Kemono Chen
Dec 12 '18 at 3:56
$begingroup$
@Kemono Chen I was just trying to give the general idea with using only angles. For more rigorous proof, your approach is correct.
$endgroup$
– Raghav
Dec 12 '18 at 10:41
add a comment |
$begingroup$
Just reposting Matt Samuel's comment as an answer:
The slope is the tangent of the angle. The tangent is a nonlinear function (so going from, for example, 5 to 25 degrees will yield a different change in slope from going from 10-30 degrees, even though the change in degrees is the same)
answered Dec 17 '18 at 21:29
user106860user106860
373315
$endgroup$
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
$begingroup$
Slope is $tan x$ , $x$ being the angle the line makes with $x$ axis. $x$ is 45 in the line $y=x$ and around 26 when $2y=x$. As you know that $tan$ approaches infinity when $x$ approaches 90, so as we add 20 to 45, the value of $tan$ will increase at a greater rate than when we add 20 to ~26.
answered Dec 12 '18 at 2:14
RaghavRaghav
647
$endgroup$
$begingroup$
No, $tan$ approaches $infty$ at $fracpi2$ is not sufficient to prove the increase is greater. You need to prove $tan''(x)>0$.
$endgroup$
– Kemono Chen
Dec 12 '18 at 3:56
$begingroup$
@Kemono Chen I was just trying to give the general idea with using only angles. For more rigorous proof, your approach is correct.
$endgroup$
– Raghav
Dec 12 '18 at 10:41
add a comment |
$begingroup$
Slope is $tan x$ , $x$ being the angle the line makes with $x$ axis. $x$ is 45 in the line $y=x$ and around 26 when $2y=x$. As you know that $tan$ approaches infinity when $x$ approaches 90, so as we add 20 to 45, the value of $tan$ will increase at a greater rate than when we add 20 to ~26.
answered Dec 12 '18 at 2:14
RaghavRaghav
647
$endgroup$
$begingroup$
No, $tan$ approaches $infty$ at $fracpi2$ is not sufficient to prove the increase is greater. You need to prove $tan''(x)>0$.
$endgroup$
– Kemono Chen
Dec 12 '18 at 3:56
$begingroup$
@Kemono Chen I was just trying to give the general idea with using only angles. For more rigorous proof, your approach is correct.
$endgroup$
– Raghav
Dec 12 '18 at 10:41
add a comment |
$begingroup$
Slope is $tan x$ , $x$ being the angle the line makes with $x$ axis. $x$ is 45 in the line $y=x$ and around 26 when $2y=x$. As you know that $tan$ approaches infinity when $x$ approaches 90, so as we add 20 to 45, the value of $tan$ will increase at a greater rate than when we add 20 to ~26.
answered Dec 12 '18 at 2:14
RaghavRaghav
647
$endgroup$
Slope is $tan x$ , $x$ being the angle the line makes with $x$ axis. $x$ is 45 in the line $y=x$ and around 26 when $2y=x$. As you know that $tan$ approaches infinity when $x$ approaches 90, so as we add 20 to 45, the value of $tan$ will increase at a greater rate than when we add 20 to ~26.
answered Dec 12 '18 at 2:14
RaghavRaghav
647
answered Dec 12 '18 at 2:14
RaghavRaghav
647
answered Dec 12 '18 at 2:14
RaghavRaghav
647
answered Dec 12 '18 at 2:14
RaghavRaghav
647
647
$begingroup$
No, $tan$ approaches $infty$ at $fracpi2$ is not sufficient to prove the increase is greater. You need to prove $tan''(x)>0$.
$endgroup$
– Kemono Chen
Dec 12 '18 at 3:56
$begingroup$
@Kemono Chen I was just trying to give the general idea with using only angles. For more rigorous proof, your approach is correct.
$endgroup$
– Raghav
Dec 12 '18 at 10:41
add a comment |
$begingroup$
No, $tan$ approaches $infty$ at $fracpi2$ is not sufficient to prove the increase is greater. You need to prove $tan''(x)>0$.
$endgroup$
– Kemono Chen
Dec 12 '18 at 3:56
$begingroup$
@Kemono Chen I was just trying to give the general idea with using only angles. For more rigorous proof, your approach is correct.
$endgroup$
– Raghav
Dec 12 '18 at 10:41
$begingroup$
No, $tan$ approaches $infty$ at $fracpi2$ is not sufficient to prove the increase is greater. You need to prove $tan''(x)>0$.
$endgroup$
– Kemono Chen
Dec 12 '18 at 3:56
$begingroup$
No, $tan$ approaches $infty$ at $fracpi2$ is not sufficient to prove the increase is greater. You need to prove $tan''(x)>0$.
$endgroup$
– Kemono Chen
Dec 12 '18 at 3:56
$begingroup$
@Kemono Chen I was just trying to give the general idea with using only angles. For more rigorous proof, your approach is correct.
$endgroup$
– Raghav
Dec 12 '18 at 10:41
$begingroup$
@Kemono Chen I was just trying to give the general idea with using only angles. For more rigorous proof, your approach is correct.
$endgroup$
– Raghav
Dec 12 '18 at 10:41
add a comment |
$begingroup$
Just reposting Matt Samuel's comment as an answer:
The slope is the tangent of the angle. The tangent is a nonlinear function (so going from, for example, 5 to 25 degrees will yield a different change in slope from going from 10-30 degrees, even though the change in degrees is the same)
answered Dec 17 '18 at 21:29
user106860user106860
373315
$endgroup$
add a comment |
$begingroup$
Just reposting Matt Samuel's comment as an answer:
The slope is the tangent of the angle. The tangent is a nonlinear function (so going from, for example, 5 to 25 degrees will yield a different change in slope from going from 10-30 degrees, even though the change in degrees is the same)
answered Dec 17 '18 at 21:29
user106860user106860
373315
$endgroup$
add a comment |
$begingroup$
Just reposting Matt Samuel's comment as an answer:
The slope is the tangent of the angle. The tangent is a nonlinear function (so going from, for example, 5 to 25 degrees will yield a different change in slope from going from 10-30 degrees, even though the change in degrees is the same)
answered Dec 17 '18 at 21:29
user106860user106860
373315
$endgroup$
Just reposting Matt Samuel's comment as an answer:
The slope is the tangent of the angle. The tangent is a nonlinear function (so going from, for example, 5 to 25 degrees will yield a different change in slope from going from 10-30 degrees, even though the change in degrees is the same)
answered Dec 17 '18 at 21:29
user106860user106860
373315
answered Dec 17 '18 at 21:29
user106860user106860
373315
answered Dec 17 '18 at 21:29
user106860user106860
373315
answered Dec 17 '18 at 21:29
user106860user106860
373315
373315
add a comment |
add a comment |
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7
$begingroup$
Because the slope the tangent of the angle, and the tangent function is nonlinear.
$endgroup$
– Matt Samuel
Dec 12 '18 at 2:00
$begingroup$
@MattSamuel That helps, thank you. but now I'm wondering why the tangent function is nonlinear.
$endgroup$
– user106860
Dec 12 '18 at 2:05
2
$begingroup$
Perhaps you should wonder why it would be linear!
$endgroup$
– The Chaz 2.0
Dec 12 '18 at 2:06
$begingroup$
@user Tongue in cheek, that seems more like a philosophy question to me. From my perspective, it is what it is. But perhaps someone has a more insightful answer than that.
$endgroup$
– Matt Samuel
Dec 12 '18 at 2:07
2
$begingroup$
A rotation can't add the same constant to every slope, because then it would never make a line vertical.
$endgroup$
– Robert Israel
Dec 12 '18 at 2:18