# Why do we need implicit differentiation in this related rates problem?

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

## 1 Answer

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

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