1. ## Implicit differentiation problem

A weather balloon V, rising vertically from a point P on the ground, is observed from a point O on the ground, 90 m from P. Let θ be the angle between OP and OV (i.e. the angle between the ground and the sightline to the balloon). At what rate is the balloon rising when θ=45˚ and θ is increasing at 1˚/second?

So, I'm thinking the job is to find:

d
(VP)/dt at the time when /dt=1.

VP(
θ)=90tanθ

and

θ(VP)=arctan(VP/90)

So I take the derivative of both functions, accounting for the fact that both VP and θ are functions of time as well as of each other:

Knowing that /dt=1 and OP=90, I get that:

Which is 1/180. I notice that if I flip this over and solve for
d(VP)/dt in the other equation, I get that

d(VP)/dt=1

Is this right?? If so, should I have been able to see the from the start? A lot of steps in this one, and I feel like I'm kinda guessing as to what I should do.

2. Originally Posted by TwoPlusTwo
A weather balloon V, rising vertically from a point P on the ground, is observed from a point O on the ground, 90 m from P. Let θ be the angle between OP and OV (i.e. the angle between the ground and the sightline to the balloon). At what rate is the balloon rising when θ=45˚ and θ is increasing at 1˚/second?

So, I'm thinking the job is to find:

d
(VP)/dt at the time when /dt=1.

VP(
θ)=90tanθ

and

θ(VP)=arctan(VP/90)

So I take the derivative of both functions, accounting for the fact that both VP and θ are functions of time as well as of each other:

Knowing that /dt=1 and OP=90, I get that:

Which is 1/180. I notice that if I flip this over and solve for
d(VP)/dt in the other equation, I get that

d(VP)/dt=1

Is this right?? If so, should I have been able to see the from the start? A lot of steps in this one, and I feel like I'm kinda guessing as to what I should do.
note that the rate of change in $\theta$ has to be in radians ...

$\frac{d\theta}{dt} = \frac{\pi}{180}$ rad/sec

let $h = PV$

$\theta = \arctan\left(\frac{h}{90}\right)$

$\frac{d\theta}{dt} = \frac{\frac{1}{90}}{1 + \left(\frac{h}{90}\right)^2} \cdot \frac{dh}{dt}$

when $\theta = 45^\circ$ , $h = 90'$ ...

$\frac{\pi}{180} = \frac{1}{180} \cdot \frac{dh}{dt}$

$\frac{dh}{dt} = \pi$ ft/sec

3. Great! So how do we know the angle has to be in radians? That didn't even occur to me, since the problem was stated using degrees.

4. Originally Posted by TwoPlusTwo
Great! So how do we know the angle has to be in radians? That didn't even occur to me, since the problem was stated using degrees.
Basically, you probably have learned that

$\frac{d}{dx} \tan x = \sec^2 x$

However, this statement is only true when x is expressed in radians. The derivatives/anti-derivatives of the trig functions are actually different if you express them in degrees.

For example, if you REALLY wanted to take a derivative in degrees, you'd actually get this:

$\frac{d}{dy} \tan y° = \frac{\pi}{180} \sec^2 y°$

You could use this to solve this problem, but instead it's generally a lot easier to convert everything to radians first!