# Differentiation + Trigonometry Question

I'm having trouble with a real-life application of Trigonometry.

Question 8b.

If someone could be kind enough to take a screenshot and attach it to this thread, I'd appreciate it.

Right, so we are given the formula for Kate's velocity.

$$V = \frac{21}{24\sin \theta + 7\cos \theta}$$ It can be shown, $$24\sin \theta + 7\cos \theta = 25\cos(\theta -73.74)$$ $$V = \frac{21}{25\cos(\theta - 73.74)}$$

The question then states: Assuming $0<\theta<150$ find the minimum value of $V$

My first attempt at this question, I did the following: The minimum of $\cos (f(\theta)) = -1$ therefore, the minimum of $V = -\frac{21}{25}$ but it says that the minimum of $V = \frac{21}{25}$

• Quick work, I like it, thanks @mrnovice – ddsffsdsdfff Feb 26 '17 at 20:48

Note that $V=\frac{21}{25\cos(\theta-73.74^{\circ})}$ is never negative on the interval $0< \theta < 150^{\circ}$ as shown on the graph below. Therefore, to minimize $V$, $\cos(\theta-73.74^{\circ})=1$.
Therefore, the minimum value is $V=\frac{21}{25}$.
• I'm silly, thanks. So, assuming $V$ could be negative then the minimum would be $-21/25$ right – ddsffsdsdfff Feb 26 '17 at 20:58
• Actually that is not true. The minimum value for the full domain would be $-\infty$ since $\cos(f(\theta))$ can get very close to zero (Which means that the denominator will get very close to zero too), and thus the whole expression would tend to $-\infty$ as you can see on the graph. – projectilemotion Feb 26 '17 at 21:01
• Was going to post exact same thing, but you beat me to it. Also worth nothing that you can find out that V is positive within the range of $\theta$ provided by seeing when $\cos(\theta-73.74) = 0$ to identify the asmyptotes. If V could be negative then the problem changes. Though in the context of the problem of crossing a road this wouldn't make much sense. So in an exam, you could use the context to conclude your answer could be wrong and then investigate further. – mrnovice Feb 26 '17 at 21:02