IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

camtorhater

A kite, 60 m above the ground, is moving horizontally at the rate of 5m/s. At what rate is the inclination of the string to the horizontal diminishing when 120 m of the string are paid out?

krish

The vertical height from ground of the kite is given to be 60m

Let the horizontal distance of the kite be x

then

dx/dt = 5 m/s (Given) -------(1)

If we draw the diagram and mark angle A between the horizontal and hypotenuse ,

we get tan A = 60/x

or x = 60 cot A

Differentiating w.r.t A

dx/dA = -60 cosec²A

Our equation no. 1 can also be written as

(dx/dA) (dA/dt) = 5 m/s

(ie) -60cosec²A (dA/dt) = 5

dA/dt = -(sin²A)/12

When string paid out is 120 m means the hypotenuse = 120 m

in this case , sin A = 60/120 = 1/2

Therefore dA/dt = -1/48 degrees per sec

Rate of diminishing = -dA/dt = 1/48 degrees per second

if there is something wrong in my soln plz do reply back.....

camtorhater

Sorry but i myself dont know the answer.

Well the experts will give the answer thats for sure.

but i do it in the following way ( plz see where am i getting the question wrong)

Suppose the horizontal distance is k

then when the string is paid out, it forms a rt triangle and

we get

=30 (sin 30=60/120)

also,

tan

=60/k

or

=tan^ -1(60/k)

now

d

/dt=d

/dk * dk/dt

=d(tan^ -1 (60/k))/dk * dk/dt

using calculator( i dont know how to get derivative of tan inverse)

i get,

answer= 1.19 degrees/sec

so where am i getting the answer wrong (i am not sure of my method, it looks a bit wrong. i dont know where but it looks so)

camtorhater

someone has done it in the following way (i had also posted on yahoo)

Related rates problem.

This is going to be tricky without using diagrams, but I'll give it a shot, and hope it is clear to you.

Draw a (right) triangle;

Horiz leg is 'x'
Vert leg is 'y'
Hypotenuse is 'r'

The angle is on the left, the vertical side is on the right.

Let the angle be '?'.

As you know: x² + y² = r²

Will come back to that.

You need a relation involving the angle. Let's try: x = rCos(?)

Take the derivative with respect to time: dx/dt = (dr/dt)Cos(?) - rSin(?)(d?/dt) ~~~~~ Eq'n (1)

From the problem we're given info....let's identify it.

Horizontal rate.....(dx/dt) = 5 m/s
Horizontal distance 'x' = 120 m
Vertical distance 'y' = 60 m

We can figure out the hypotenuse by Pythagorean Thrm. I'll let you actually do it, but 'r' = 300 m

So...now what? Well, we don't know what dr/dt is in eq'n (1).

Well, we know that dx/dt = 5 m/s, and we can use the following:

In the same proportions: (dy/dt)/(dx/dt) = y/x

==> dy/dt = (y/x)(dx/dt) = (60 m/120 m)(5 m/s) = 5/2 m/s

Now we can use is the fact that since x² + y² = r²
Then (dx/dt)² + (dy/dt)² = (dr/dt)²

The velocities must also be held to this relationship because it involves the same variables (in essence).

So, dr/dt = ?[(dx/dt)² + (dy/dt)²] = (5/2)?5 m/s

Now, we can plug this stuff into our eq'n (1)

==> dx/dt = (dr/dt)Cos(?) - rSin(?)(d?/dt) (you want the rate of change of the angle with respect to time)
==> (d?/dt) = [(dr/dt)Cos(?) - dx/dt]/[rCos(?)]

Plug your numbers in and you should see that:

(d?/dt) ? 0.118°/s

camtorhater

i think my answer is wrong bcoz some of my friends tell me that tan^ -1x is not differentiable (with respect to x). well never mind but i am eagerly waiting for a reply from the experts.

krishna.gopal

Krish every thing is right with your answer except that dA/at is -1/48 radian per sec instead of degrees per second.

And there is another way to look at this problem

dA/dt=

=(v tangential)/r (r is the radius of kite from man holding it ie. 120 m)

V tangential is equall to 5sin(30)=2.5m/s

Therefore dA/at=1/48 (in magnitude)

camtorhater

sir, thank you for answering the question

you have written ans is -1/48 rad per sec

which i think is equal to -1.19 degrees/sec(which is my answer). so is my procedure correct or not

krishna.gopal

Friend your answer is very correct. Nothing wrong in doing it that way. What i was trying to put was the fact that a bit of application of physics here makes it a one liner

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