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A hot air balloon leaves the ground at a point that is horizontal distance 190 metres from an observer and rises vertically upwards. The observer notes that the rate of increase of the angle of inclination between the horizontal and the vertical and the observer where line of sight (hyp) is a constant 0.045 radians per minute.

What is the rate of increase in metres per minute of the distance between the balloon and the observer when the balloon is 180 metres above the ground?

To solve it I visualised a right angled triangle with angle 0.045 radians, a base (adj) of 190 and opposite of 180m (position of balloon) so I'm guessing an element of trig is involved. I denoted the rate of increase in metres per minute of the distance ds/dt but I don't know if s is the hypotenuse and what to do to calculate ds/dt.

Can someone please break it down for me?


i did but i hit a brick wall. If i call the side opposite to θ x then tanθ=x/170 and dx/dθ=170sec$^2$θ (which can be expressed as cos) but I don't know what to do next

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I think using the tangent is the most straightforward way. Write an equation using $tan(\theta)$ and then take the derivative of both sides with respect to time. Then plug for what you know and solve for the rate of change of the opposite side. – Todd Wilcox Oct 31 '12 at 16:12
I thought that and got dx/dt=170sec$^2$dθ/dt (don't know if that's right) but thwen you don't make use of the 200metres given in the question – Ricky Rozay Oct 31 '12 at 16:20
up vote 2 down vote accepted

Let $y$ represent the height above ground at any moment, $x$ represent the horizontal distance from the observer to the launch point, $h$ represent the distance from the observer to the balloon, and $\theta$ represent the angle above the horizontal at which the observer sees the balloon at any moment. Then,$$\sec(\theta)=\dfrac{h}{x}$$Since $x$ is fixed we can re-write that as,$$\sec(\theta)=\dfrac{h}{190}$$ Taking the time derivative of both sides we get,$$\sec(\theta)\tan(\theta)\frac{d\theta}{dt}=\frac{1}{190}\frac{dh}{dt}$$ Now find $\theta$ using the instantaneous height and fixed distance given, plug in $\theta$ and $\frac{d\theta}{dt}$ and solve for $\frac{dh}{dt}$.

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perhaps express the elevation of balloon as a function of $\cos \theta$, where $\theta$ the angle to the observer. Taking derivatives in time will get you to speed of elevation on one side and $\sin \theta \cdot \frac{d \theta}{dt}$ on the other side. Note that you have $d \theta/dt$.

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