I'm differentiating this wrong!
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I am self-teaching calculus and have been looking at the related rates practice problems here: https://www.whitman.edu/mathematics/calculus_online/section06.02.html
I am having trouble with the last problem, which relates to the rate at which a piece of paper will be cut by a pair of scissors. We are given the rate $dot{theta}$ at which the angle of a pair of open scissors ($theta$) changes with respect to time ($t$) and also some other numbers based on the diagram provided and a fixed point when the scissors are closing.
I'm happy that $x = (alpha sin beta)cdot sin(beta+theta)^{-1}$ as stated in the provided solution, which I arrived at by the law of sines. However, when I implicitly differentiate this and plug in the provided values, my answer is way off.
The task is to find the derivative of $x$ with respect to time, $dot{x}$. $alpha$ and $beta$ are constants, 20cm and 5 degrees respectively. $theta$ varies, but we know that its derivative with respect to time, $dot{theta}$, is -50 degrees per second.
My approach to differentiation has been to start off with the angle addition identity for $sin(a+b)$ to say that
$$dot{x} = (alpha sinbeta) cdot frac{d}{dt} sin(beta+theta)^{-1} = (alpha sinbeta) cdot frac{d}{dt} (sin(theta)cos(beta)+cos(theta)sin(beta))^{-1}$$
I'm then using the chain rule to find $frac{d}{dt}(sin(theta)cos(beta)+cos(theta)sin(beta))^{-1}$, and I'm ending up with:
$$-1cdot(sin(theta)cos(beta)+cos(theta)sin(beta))^{-2}cdot(cos(beta)cos(theta)-sin(theta)sin(beta))cdotdot{theta}$$
I then simplify this a bit, but I suspect I have done it wrong as plugging in the provided values doesn't get me to the right answer (~$3.79$cm/s)!
Can you see what I'm doing wrong? Any help is greatly appreciated.
implicit-differentiation
edited Dec 22 '18 at 21:05
postmortes
2,07531120
asked Dec 22 '18 at 11:52
Nicholas James BaileyNicholas James Bailey
1483
$endgroup$
add a comment |
$begingroup$
I am self-teaching calculus and have been looking at the related rates practice problems here: https://www.whitman.edu/mathematics/calculus_online/section06.02.html
I am having trouble with the last problem, which relates to the rate at which a piece of paper will be cut by a pair of scissors. We are given the rate $dot{theta}$ at which the angle of a pair of open scissors ($theta$) changes with respect to time ($t$) and also some other numbers based on the diagram provided and a fixed point when the scissors are closing.
I'm happy that $x = (alpha sin beta)cdot sin(beta+theta)^{-1}$ as stated in the provided solution, which I arrived at by the law of sines. However, when I implicitly differentiate this and plug in the provided values, my answer is way off.
The task is to find the derivative of $x$ with respect to time, $dot{x}$. $alpha$ and $beta$ are constants, 20cm and 5 degrees respectively. $theta$ varies, but we know that its derivative with respect to time, $dot{theta}$, is -50 degrees per second.
My approach to differentiation has been to start off with the angle addition identity for $sin(a+b)$ to say that
$$dot{x} = (alpha sinbeta) cdot frac{d}{dt} sin(beta+theta)^{-1} = (alpha sinbeta) cdot frac{d}{dt} (sin(theta)cos(beta)+cos(theta)sin(beta))^{-1}$$
I'm then using the chain rule to find $frac{d}{dt}(sin(theta)cos(beta)+cos(theta)sin(beta))^{-1}$, and I'm ending up with:
$$-1cdot(sin(theta)cos(beta)+cos(theta)sin(beta))^{-2}cdot(cos(beta)cos(theta)-sin(theta)sin(beta))cdotdot{theta}$$
I then simplify this a bit, but I suspect I have done it wrong as plugging in the provided values doesn't get me to the right answer (~$3.79$cm/s)!
Can you see what I'm doing wrong? Any help is greatly appreciated.
implicit-differentiation
edited Dec 22 '18 at 21:05
postmortes
2,07531120
asked Dec 22 '18 at 11:52
Nicholas James BaileyNicholas James Bailey
1483
$endgroup$
$begingroup$
Does $sin(beta + theta)^{-1}$ means $sinfrac{1}{beta + theta}$ or $frac{1}{sin(beta + theta)}$?
$endgroup$
– Alex Vong
Dec 22 '18 at 21:48
$begingroup$
From the link it seems to be the latter, $$frac{1}{sin(beta+theta)}$$
$endgroup$
– mark
Dec 22 '18 at 22:17
$begingroup$
Indeed - the latter! Any ideas what's going on? Thanks for your help.
$endgroup$
– Nicholas James Bailey
Dec 22 '18 at 23:44
add a comment |
$begingroup$
I am self-teaching calculus and have been looking at the related rates practice problems here: https://www.whitman.edu/mathematics/calculus_online/section06.02.html
I am having trouble with the last problem, which relates to the rate at which a piece of paper will be cut by a pair of scissors. We are given the rate $dot{theta}$ at which the angle of a pair of open scissors ($theta$) changes with respect to time ($t$) and also some other numbers based on the diagram provided and a fixed point when the scissors are closing.
I'm happy that $x = (alpha sin beta)cdot sin(beta+theta)^{-1}$ as stated in the provided solution, which I arrived at by the law of sines. However, when I implicitly differentiate this and plug in the provided values, my answer is way off.
The task is to find the derivative of $x$ with respect to time, $dot{x}$. $alpha$ and $beta$ are constants, 20cm and 5 degrees respectively. $theta$ varies, but we know that its derivative with respect to time, $dot{theta}$, is -50 degrees per second.
My approach to differentiation has been to start off with the angle addition identity for $sin(a+b)$ to say that
$$dot{x} = (alpha sinbeta) cdot frac{d}{dt} sin(beta+theta)^{-1} = (alpha sinbeta) cdot frac{d}{dt} (sin(theta)cos(beta)+cos(theta)sin(beta))^{-1}$$
I'm then using the chain rule to find $frac{d}{dt}(sin(theta)cos(beta)+cos(theta)sin(beta))^{-1}$, and I'm ending up with:
$$-1cdot(sin(theta)cos(beta)+cos(theta)sin(beta))^{-2}cdot(cos(beta)cos(theta)-sin(theta)sin(beta))cdotdot{theta}$$
I then simplify this a bit, but I suspect I have done it wrong as plugging in the provided values doesn't get me to the right answer (~$3.79$cm/s)!
Can you see what I'm doing wrong? Any help is greatly appreciated.
implicit-differentiation
edited Dec 22 '18 at 21:05
postmortes
2,07531120
asked Dec 22 '18 at 11:52
Nicholas James BaileyNicholas James Bailey
1483
$endgroup$
I am self-teaching calculus and have been looking at the related rates practice problems here: https://www.whitman.edu/mathematics/calculus_online/section06.02.html
I am having trouble with the last problem, which relates to the rate at which a piece of paper will be cut by a pair of scissors. We are given the rate $dot{theta}$ at which the angle of a pair of open scissors ($theta$) changes with respect to time ($t$) and also some other numbers based on the diagram provided and a fixed point when the scissors are closing.
I'm happy that $x = (alpha sin beta)cdot sin(beta+theta)^{-1}$ as stated in the provided solution, which I arrived at by the law of sines. However, when I implicitly differentiate this and plug in the provided values, my answer is way off.
The task is to find the derivative of $x$ with respect to time, $dot{x}$. $alpha$ and $beta$ are constants, 20cm and 5 degrees respectively. $theta$ varies, but we know that its derivative with respect to time, $dot{theta}$, is -50 degrees per second.
My approach to differentiation has been to start off with the angle addition identity for $sin(a+b)$ to say that
$$dot{x} = (alpha sinbeta) cdot frac{d}{dt} sin(beta+theta)^{-1} = (alpha sinbeta) cdot frac{d}{dt} (sin(theta)cos(beta)+cos(theta)sin(beta))^{-1}$$
I'm then using the chain rule to find $frac{d}{dt}(sin(theta)cos(beta)+cos(theta)sin(beta))^{-1}$, and I'm ending up with:
$$-1cdot(sin(theta)cos(beta)+cos(theta)sin(beta))^{-2}cdot(cos(beta)cos(theta)-sin(theta)sin(beta))cdotdot{theta}$$
I then simplify this a bit, but I suspect I have done it wrong as plugging in the provided values doesn't get me to the right answer (~$3.79$cm/s)!
Can you see what I'm doing wrong? Any help is greatly appreciated.
implicit-differentiation
implicit-differentiation
edited Dec 22 '18 at 21:05
postmortes
2,07531120
asked Dec 22 '18 at 11:52
Nicholas James BaileyNicholas James Bailey
1483
edited Dec 22 '18 at 21:05
postmortes
2,07531120
asked Dec 22 '18 at 11:52
Nicholas James BaileyNicholas James Bailey
1483
edited Dec 22 '18 at 21:05
postmortes
2,07531120
edited Dec 22 '18 at 21:05
postmortes
2,07531120
edited Dec 22 '18 at 21:05
postmortes
2,07531120
2,07531120
asked Dec 22 '18 at 11:52
Nicholas James BaileyNicholas James Bailey
1483
asked Dec 22 '18 at 11:52
Nicholas James BaileyNicholas James Bailey
1483
asked Dec 22 '18 at 11:52
Nicholas James BaileyNicholas James Bailey
1483
1483
$begingroup$
Does $sin(beta + theta)^{-1}$ means $sinfrac{1}{beta + theta}$ or $frac{1}{sin(beta + theta)}$?
$endgroup$
– Alex Vong
Dec 22 '18 at 21:48
$begingroup$
From the link it seems to be the latter, $$frac{1}{sin(beta+theta)}$$
$endgroup$
– mark
Dec 22 '18 at 22:17
$begingroup$
Indeed - the latter! Any ideas what's going on? Thanks for your help.
$endgroup$
– Nicholas James Bailey
Dec 22 '18 at 23:44
add a comment |
$begingroup$
Does $sin(beta + theta)^{-1}$ means $sinfrac{1}{beta + theta}$ or $frac{1}{sin(beta + theta)}$?
$endgroup$
– Alex Vong
Dec 22 '18 at 21:48
$begingroup$
From the link it seems to be the latter, $$frac{1}{sin(beta+theta)}$$
$endgroup$
– mark
Dec 22 '18 at 22:17
$begingroup$
Indeed - the latter! Any ideas what's going on? Thanks for your help.
$endgroup$
– Nicholas James Bailey
Dec 22 '18 at 23:44
$begingroup$
Does $sin(beta + theta)^{-1}$ means $sinfrac{1}{beta + theta}$ or $frac{1}{sin(beta + theta)}$?
$endgroup$
– Alex Vong
Dec 22 '18 at 21:48
$begingroup$
Does $sin(beta + theta)^{-1}$ means $sinfrac{1}{beta + theta}$ or $frac{1}{sin(beta + theta)}$?
$endgroup$
– Alex Vong
Dec 22 '18 at 21:48
$begingroup$
From the link it seems to be the latter, $$frac{1}{sin(beta+theta)}$$
$endgroup$
– mark
Dec 22 '18 at 22:17
$begingroup$
From the link it seems to be the latter, $$frac{1}{sin(beta+theta)}$$
$endgroup$
– mark
Dec 22 '18 at 22:17
$begingroup$
Indeed - the latter! Any ideas what's going on? Thanks for your help.
$endgroup$
– Nicholas James Bailey
Dec 22 '18 at 23:44
$begingroup$
Indeed - the latter! Any ideas what's going on? Thanks for your help.
$endgroup$
– Nicholas James Bailey
Dec 22 '18 at 23:44
add a comment |
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$begingroup$
Does $sin(beta + theta)^{-1}$ means $sinfrac{1}{beta + theta}$ or $frac{1}{sin(beta + theta)}$?
$endgroup$
– Alex Vong
Dec 22 '18 at 21:48
$begingroup$
From the link it seems to be the latter, $$frac{1}{sin(beta+theta)}$$
$endgroup$
– mark
Dec 22 '18 at 22:17
$begingroup$
Indeed - the latter! Any ideas what's going on? Thanks for your help.
$endgroup$
– Nicholas James Bailey
Dec 22 '18 at 23:44