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I have a question about number 3. I understand how to do the problem but the answer to this used cotx. I used tanx and got the wrong answer. Can someone explain why they used cotangent

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  • $\begingroup$ Cotangent is just the inverse of tangent, so you probably got the fraction upside down. Without seeing your work we can't see what was wrong. There is nothing magic about using one or the other as long as you use it properly. $\endgroup$ – Ross Millikan Dec 14 '16 at 2:57
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Diagram of situation

Since tangent is opposite over adjacent, and the angle is adjacent to 8t m leg and opposite to the 100 m leg, $\tan{\theta}=\frac{100}{8t}$ and therefore $\theta = \tan^{-1}{\frac{100}{8t}}$. Differentiating this gives the answer of $$\frac{d\theta}{dt} = \frac{100}{8(1+(\frac{100}{8t})^2)} $$

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