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There is this question on derivatives to which the answer is given as $\frac{43}{75}$rad/sec in the answers section. This answer appears to be wrong.

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My Solution $$ \theta+\tan^{-1}\left(\frac{6}{4-x}\right)+\tan^{-1}\left(\frac{3}{x}\right)=\pi $$ which gets reduced to $$ \theta=\pi-\tan^{-1}\left(\frac{3x+12}{4x-x^2-18}\right) $$

Therefore, $$ \frac{d\theta}{dt}=\left(\frac{3x^2+24x-102}{(4x-x^2+18)^2+(3x+12)^2}\right)\frac{dx}{dt} $$

Given $x=4$ and $\frac{dx}{dt}=2\text{ cm/sec}$,

$$ \frac{d\theta}{dt}=-\frac{7}{75}\text{rad/sec} $$

I've got the graph of $\theta$ as a function of $x$ here, which also indicates my answer is correct. Or am I? Kindly help.

Abhishek A Udupa
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    How did you add the two arctan expressions? You can’t combine them together in one tan^(-1) – IrbidMath Aug 28 '21 at 15:23
  • You would also get the same $-\frac{7}{75}$ is you said $\theta=\pi-\tan^{-1}\left(\frac{6}{4-2t}\right)-\tan^{-1}\left(\frac{3}{2t}\right)$ and $\frac{d\theta}{dt}=-\frac{3}{{{t}^{2}}-4 t+13}+\frac{6}{4 {{t}^{2}}+9}$ and then let $t=2$ so your arithmetic is correct – Henry Aug 28 '21 at 15:24
  • @IrbidMath: There is an arctangent addition formula, though I agree it seems to be an unnecessary complication to use it here. – Troposphere Aug 28 '21 at 15:24
  • $\theta = \pi\mathbf{+} \arctan(\dots)$. (Test at $x = 4/3$, for instance.) – Eric Towers Aug 28 '21 at 15:25
  • You can only use the arctangent addition formula $\frac{6}{4-x}\cdot \frac{3}{x}<1$. This happens when $x<0$ or $x>4$ which doesn't apply in this problem. Personally, I would use the Law of Cosines to measure $\theta$. – Matthew H. Aug 28 '21 at 15:53

3 Answers3

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Here's an alternative explicit calculation.

Let $D$ be the end of the $6$-cm segment opposite $C.$ Let $E$ be the end of the $3$-cm segment opposite $A.$

The angle $\theta$ is the sum of the angles $\angle BDC$ and $\angle BEA.$ We can see that when $x=4,$ then $B$ has just reached $C$ and is moving perpendicular to the $6$ cm segment $BC$ at the rate $2$ cm per second, so the angular velocity of $B$ around the point $D$ is $2/6 = \frac13$ cm per second clockwise and therefore $\angle BDC$ is changing at the rate $-\frac13$ radian per second.

Also when $x = 4,$ the distance from $B$ to $E$ is $5$ cm. Consider the radial and tangential components of the velocity vector of $B$ relative to the circle of radius $5$ around $E.$ The velocity vector and its components make a $3,4,5$ right triangle with the magnitude of the tangential component equal to $\frac35$ the magnitude of the velocity; that is, the tangential component is $\frac35\cdot 2 = \frac65$ cm per second clockwise. This means that the angle $\angle BEA$ is increasing at the rate $\frac{6/5}{5} = \frac{6}{25}$ cm per second.

So the answer is $$ -\frac13 + \frac{6}{25} = - \frac{25}{75} + \frac{18}{25} = - \frac{7}{75}, $$ exactly as you found.

Note that $$ \frac{25}{75} + \frac{18}{25} = \frac{43}{75}, $$ so the supposed answer in the book appears to be the result of a sign error.

David K
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The book's answer of $\frac{43}{75}$ rad/s obviously can't be true.

When $x=4$ point $B$ coincides with $C$, and then the angle above $\theta$ on the diagram (that is, between the upper dashed line and $BC$) changes by exactly $2/6 =\frac13$ rad/s. The angle below $\theta$ is changing in the opposite direction (and by no more than $\frac25$ rad/s, since the distance to the lower right corner is $5$), so the absolute value of the correct answer is at most $\frac 13$.

However $\frac{43}{75}$ is clearly larger than $\frac13=\frac{25}{75}$.

Since commenters have confirmed your result with a different calculation, it looks very likely that it is correct.

Troposphere
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  • With regard to the absolute value of $d\theta/dt,$ it also matters how fast the angle below $\theta$ is changing. If that angle had an extremely large negative rate of change then the absolute value of $d\theta/dt$ could be greater than $1.$ But in fact the other angle's rate of change has magnitude less than $\frac23$, so indeed $|d\theta/dt|\leq\frac13.$ – David K Aug 28 '21 at 15:47
  • @DavidK: True, that's a possibility I hadn't explicitly excluded. Edited. – Troposphere Aug 28 '21 at 16:06
  • @Troposphere The answer given in the text book is $\frac{43}{75} \text{ rad/sec}$ and not $\frac{43}{25} \text{ rad/sec}$. I apologise for the mistake and has been rectified in the question asked. – Abhishek A Udupa Aug 28 '21 at 16:10
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Differentiating your formula $$\theta+\arctan\frac{6}{4-x}+\arctan\frac{3}{x}=\pi$$ with respect to time gives $$\dot\theta+\frac1{1+\left(\dfrac6{4-x}\right)^2}\cdot\frac6{({4-x})^2}\dot x+\frac1{1+\left(\dfrac3x\right)^2}\cdot\frac{-3}{x^2}\dot x=0.$$ After substituting $\dot x=2$, this may be written $$\dot\theta=6\left(\frac1{x^2+9}-\frac2{(4-x)^2+36}\right).$$ Now substituting $x=4$ yields the answer $\dot\theta=-\frac7{75}$.

John Bentin
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