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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

$$\dot{x} = (\alpha \sin\beta) \cdot \frac{d}{dt} \sin(\beta+\theta)^{-1} = (\alpha \sin\beta) \cdot \frac{d}{dt} (\sin(\theta)\cos(\beta)+\cos(\theta)\sin(\beta))^{-1}$$

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:

$$-1\cdot(\sin(\theta)\cos(\beta)+\cos(\theta)\sin(\beta))^{-2}\cdot(\cos(\beta)\cos(\theta)-\sin(\theta)\sin(\beta))\cdot\dot{\theta}$$

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.

postmortes
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Nicholas James Bailey
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  • Does $\sin(\beta + \theta)^{-1}$ means $\sin\frac{1}{\beta + \theta}$ or $\frac{1}{\sin(\beta + \theta)}$? – Alex Vong Dec 22 '18 at 21:48
  • From the link it seems to be the latter, $$\frac{1}{\sin(\beta+\theta)}$$ – mark Dec 22 '18 at 22:17
  • Indeed - the latter! Any ideas what's going on? Thanks for your help. – Nicholas James Bailey Dec 22 '18 at 23:44
  • I don't get the result either but I think not expanding the sum makes the expression easier, $\dot{x}=\alpha\sin\beta\frac{d}{dt}\sin(\beta+\theta)^{-1} =-\alpha \sin\beta\frac{\cos(\beta+\theta)}{\sin(\beta+\theta)^2}\dot{\theta}=-20[cm]\sin5°\frac{\cos(5°+30°)}{\sin(5°+30°)^2}(-50/360)[s^{-1}]=\frac{25}{9}\frac{\cos(35°)}{\sin(35°)^2}=0.6$ – Dabed Mar 04 '20 at 23:32

1 Answers1

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You already got the first part: $x = \frac{a \sin \beta}{\sin(\theta + \beta)}$

$\frac{dx}{dt} $

$= \left(a \sin \beta \right) \frac{d}{dt} \left(\frac{1}{\sin (\theta + \beta)} \right)$

$ = \left(a \sin \beta \right) \left( \frac{-1}{\sin^2 (\theta + \beta)}\right) \cos (\theta + \beta) \left(\frac{d \theta}{dt}\right)$

Next put

$a = 20$

$\beta = 5^{\circ}$

$\theta = 30 ^{\circ}$

$\frac{d \theta}{dt} = -\frac{50^{\circ}}{180^{\circ}} \pi$ (Note the conversion from degree to radian)

Finally $\frac{dx}{dt} = 20 \frac{\sin 5^{\circ} \cos 35^{\circ}} {\sin^2 35^{\circ}} \frac{50 \pi}{180}$ cm/sec

$= 3.788$ cm/sec

PTDS
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