# Thread: writing equations for rate of change questions

1. ## writing equations for rate of change questions

Hi
Im trying to write an equation for the question below. i did write one but am pretty sure it is wrong because I need to differentiate arctan in it and we have not been taught that yet. Could someone please point me in the right direction with writing it?

A telescope is 75m above water level on a cliff and a boat is approaching at 6m/s. what is the rate of change of angle of the telescope when the boat is 75m from shore.

I thought it might be y=arctan(75/x) because the angle specified is found using arctan(opposite/addjacent).

Any suggestions appreciated

2. Originally Posted by Poppy
Hi
Im trying to write an equation for the question below. i did write one but am pretty sure it is wrong because I need to differentiate arctan in it and we have not been taught that yet. Could someone please point me in the right direction with writing it?

A telescope is 75m above water level on a cliff and a boat is approaching at 6m/s. what is the rate of change of angle of the telescope when the boat is 75m from shore.

I thought it might be y=arctan(75/x) because the angle specified is found using arctan(opposite/addjacent).

Any suggestions appreciated
HI

I got 0.04 degree per second , seems weird .

Draw a triangle , with the vertical side being 75 , horizontal = x and the angle between the hypothenus and vertical side is $\theta$

Given dx/dt=6 , $\frac{d\theta}{dt}=??$

$\frac{d\theta}{dt}=\frac{d\theta}{dx}\cdot \frac{dx}{dt}$

$=\frac{1}{150}\cdot 6=0.04$

$\tan \theta=\frac{x}{75}$

$x=75\tan \theta$

$\frac{dx}{d\theta}=75\sec^2 \theta$

so when x=75 , $\tan \theta=1\Rightarrow \theta=45^o$

$=\frac{75}{\cos^2 45^o}$

$=\frac{75}{(\frac{1}{\sqrt{2}})^2}=150$

Tell me if it doesn't match the ans in ur book .

$=\frac{75}{\cos^2 45^o}$

$=\frac{75}{(\frac{1}{\sqrt{2}})^2}=150$

Hi again, that makes sense, however, i was wondering how sec squared (45) becomes (1/√2)^2
Thanks

4. [QUOTE=Poppy;423652]
HI

$=\frac{75}{\cos^2 45^o}$

$=\frac{75}{(\frac{1}{\sqrt{2}})^2}=150$

Hi again, that makes sense, however, i was wondering how sec squared (45) becomes (1/√2)^2
Thanks
$(sec(x))^n = sec^n(x) = \frac{1}{cos^n(x)} \: \: , \: n \in \mathbb{Z^+}$

$cos(45) = \frac{1}{\sqrt2}$ (This is a special angle-see spoiler for why)

Spoiler:
To show this construct an isosceles right angled triangle with the short sides being equal to 1. By definition the angles will be 45 degrees. Using Pythagoras Theorem we see the hypotenuse is $\sqrt2$. You can then use the basic ratio of cos(45)

$sec^2(x) = \frac{1}{cos^2(x)}$ and so we get

$\frac{1}{\left(\frac{1}{\sqrt2}\right)^2} = 2$