Question

a balloon rising straight up from level ground is tracked by a man with a video camera, 200 meters from the point of lift off. at the moment the video camera's angle of elevation is pi/6, the angle is increasing at the rate of 0.21 rad/min. how fast is the balloon rising at that moment (neglect the height of the camera)?..

Answer 1

First, let's work out the value of a

tanθ = o/a
tanPi/6 = 200/a
1/rt(3) = 200/a
a = 200rt(3)

\[a=200\sqrt{3}\]

Answer 2

i think the length 200 meters is the length adjacent to the angle..

Answer 3

Haha, good observation. I will fix it

Answer 4

its a differential calculus problem.. and im poor at differentiating...so if u can figure out this one.. your my hero... haha...

Answer 5

So we have:

tanθ=o/a
tanθ=x/200
θ=tan^-1(x/200)

Now when we differentiate this we get dθ/dx

Do you get stuck at working out what to differentiate or how to actually differentiate the equations?

Answer 6

both..hehe

Answer 7

Haha, okay. (I will start the question again and explain what I am doing)

What helps me when I do these questions is to write everything that you have as a derivative fraction.

In the question you are given a rate is radians/minute.
So as a fraction that is dθ/dt
Therefore:
\[\frac{d \theta}{dt}=0.21\]

Now in the actual question we are asked to find the rate of change of the balloon's height with respect to time.
As a fraction that is dx/dt (if x is taken as the distance)
\[\frac{d x}{dt}=?\]

Following so far?

Answer 8

yes then the next is?

Answer 9

can you differentiate tan^-1(x/200) for me.. as an example....

Answer 10

Not quite, I will show you.

So we have \[\frac{d \theta}{dt}\]
and we want to find,
\[\frac{d x}{dt}\]

So we need to value of \[\frac{dx}{d \theta}\] to cancel out the theta on the top of the first derivative. Like this:
\[\frac{dx}{dt} = \frac{d \theta}{dt} \times \frac{dx}{d \theta}\]

Answer 11

I will do the differentiation to find \[\frac{dx}{d\theta}\]

Answer 12

tanθ=x/200
x =200tanθ

From an identity, the derivative of tan(x) is 1/(cos(x))^2

So
\[\frac{dx}{d \theta}=200 \times \frac{1}{\cos^2 \theta}\]
\[\frac{dx}{d \theta}= \frac{200}{\cos^2\theta}\]

Answer 13

Do you know what the next step is?

Answer 14

i thought the derivative of tan(x) is sec^2(x)

Answer 15

The definition of the sec(x) function is 1/cos(x)

Take a look at sec(A) on this site.
http://en.wikipedia.org/wiki/Trigonometric_functions#Reciprocal_functions

Answer 16

yea i got it now.. i dont know whats the next,. how to fiind the rate of rising...

Answer 17

So, using the formula that I found ages ago:
\[\frac{dx}{dt}=\frac{d\theta}{dt} \times \frac{dx}{d \theta}\]

We can substitute in what we know:
\[\frac{dx}{dt}=0.21 \times 200sec^2{\theta}\]
And from the question, the value of theta is Pi/6

\[\frac{dx}{dt}=42sec^2{\frac{\pi}{6}}\]
Think you can take it from here?

Answer 18

is it possible to have an exact value without an identities

Answer 19

Yes, I was wondering if you knew how to do this but I will do it for you.
The only way to get exact values is to remember them so I use this online calculator to get them if I don't know them:

http://www.wolframalpha.com/input/?i=sec%28Pi%2F6%29

So we car rewrite the equation
\[\frac{dx}{dt}=42\frac{2}{\sqrt{3}}^2\]
\[\frac{dx}{dt}=42\frac{4}{3}\]
\[\frac{dx}{dt}=56\]

So the rate of change of the height is 56 metres per second

Answer 20

thanks....