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Two people are 50 feet apart. One of them starts walking north at a rate so that the angle shown in the diagram below is changing at a constant rate of 0.01 rad/min. At what rate is distance between the two people changing when radians?

enter image description here

so $\sec{\theta} = \frac{x}{50}$

and so I hear the next step is:

$\sec{\theta}\tan{\theta} \cdot \theta' = \frac{x'}{50}$

I don't get that step.

  1. Why do we need implicit differentiation here? I thought the derivative of $\sec{\theta}$ was $\sec{\theta}\tan{\theta}$
  2. Where did the right side come from?
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    $\begingroup$ you are probably differentiating wrt time $t$, so you need to apply the chain rule accordingly $\endgroup$
    – jonsno
    May 5, 2018 at 16:30

1 Answer 1

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For question 2

Note that we have

$$\frac {d \cos(\theta)}{d \theta}=-\sin(\theta)$$ but $$\frac {d \cos(\theta(t))}{dt}=\frac {d \cos(\theta(t))}{d \theta}\frac {d \theta}{dt}=-\sin(\theta(t)) \frac {d\theta }{dt}=-\sin(\theta(t)) \theta '$$

the derivative is not taken wrt $\theta$ but to t and $\theta=\theta(t)$ $$\sec{\theta} = \frac{x}{50} $$ $$\cos^{-1} (\theta)=\frac{x}{50} $$ You have the rule $(f^n)'=nf^{n-1} \times f'$ $$-1 \cos^{-2}(\theta)(-\sin (\theta)) \theta '=\frac{x'}{50}$$ $$ \cos^{-1} (\theta) \tan (\theta) \theta '=\frac{x'}{50}$$ $$ \sec (\theta) \tan (\theta) \theta '=\frac{x'}{50}$$

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  • $\begingroup$ And the right side we just use quotient rule right? $\endgroup$
    – Jwan622
    May 5, 2018 at 17:25
  • $\begingroup$ And the right side we just use quotient rule right? $\endgroup$
    – Jwan622
    May 5, 2018 at 17:25
  • $\begingroup$ yes its normal differentiation rules @Jwan622 $\endgroup$ May 5, 2018 at 17:29
  • $\begingroup$ But doesn't x change in response to t too? $\endgroup$
    – Jwan622
    May 5, 2018 at 17:29
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    $\begingroup$ Oh I seeeeeee I think $\endgroup$
    – Jwan622
    May 5, 2018 at 17:30

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