Related rates and flying airplane

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An airplane flying horizontally at 3 miles altitude and 480 mi/hr passes directly above an observer. After 30 seconds, the plane travels 4 miles horizontally, leading to a distance calculation using the Pythagorean theorem: D = √(3² + 4²). The discussion emphasizes deriving the distance function D with respect to time, resulting in the expression D = √(3² + (4T/30)²). Differentiating this expression will yield the rate of change of distance from the observer to the airplane, which is the desired solution.
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Homework Statement


An airplane flying horizontally at an altitude of 3 miles and at a speed of 480mi/hr passes directly above an observer on the ground. How fast is the distance from the observer to the airplane increasing 30 seconds later?



Homework Equations





The Attempt at a Solution


can you check if this is right:
If the plane is moving at 480 mi/hr, use 30 seconds to compute how far it traveled. (480 miles/hour)(hour/60 minutes) = 8 miles/minute.
It can be seen that 30 seconds is half a minutes, so the plane would have flown half of 8 miles. You can probably do this, but 8/2=4.
If the plane were 3 miles overhead and traveled 4 miles away, the distance can be found as √(3²+4²). Compute 3². Compute 4². Add the two together. The number should be recognized as X².


rest i don't know
 
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It's a rates problem. The vertical distance is 3mi. The horizontal distance is h(t) where h depends on time because the plane is moving. Right? They want to know what is the rate of change of sqrt((3mi)^2+h(t)^2). Take a derivative. Can you write an expression for h(t)?
 
hi !

yr you can derive a equation for the distance between observer and airplane

it will be like this
D=sqrt(3^2+(4T/30)^2)
hope you will know how to approach it ,here T means the time in seconds.

then just differentiate D w.r.t T
then u will get dD/dT
just subb T in the above derivative and u will get the value for the change of distance with time .i think that what u r looking for ...

it is great to help a sri lankan!reply if u want more help
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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