0
$\begingroup$

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?
$\endgroup$
  • 5
    $\begingroup$ you are probably differentiating wrt time $t$, so you need to apply the chain rule accordingly $\endgroup$ – samjoe May 5 '18 at 16:30
2
$\begingroup$

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 $(f^n)'=nf^{n-1} \times f'$ $$-1 \cos^{-2}(\theta)(-\sin (\theta)) \theta '=\frac{x'}{50}$$ $$ \cos^{-1} (\theta) \tan (\theta) \theta '=\frac{x'}{50}$$ $$ \sec (\theta) \tan (\theta) \theta '=\frac{x'}{50}$$

$\endgroup$
  • $\begingroup$ And the right side we just use quotient rule right? $\endgroup$ – Jwan622 May 5 '18 at 17:25
  • $\begingroup$ And the right side we just use quotient rule right? $\endgroup$ – Jwan622 May 5 '18 at 17:25
  • $\begingroup$ yes its normal differentiation rules @Jwan622 $\endgroup$ – Isham May 5 '18 at 17:29
  • $\begingroup$ But doesn't x change in response to t too? $\endgroup$ – Jwan622 May 5 '18 at 17:29
  • 1
    $\begingroup$ Oh I seeeeeee I think $\endgroup$ – Jwan622 May 5 '18 at 17:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.