word prob on derivatives?

Question

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?

Answer 1

At t=0, the horizontal distance from the vertex of the string to the point below the kite is H0
where H0=sqrt(120^2-60^2)
this is computed as
104 m

at t>0

cos(a)=(H0+5*t)/L
where a is the angle and L is the length of the string
We also nac say that at any time
L=60/sin(a)

so the cot(a)=(H0+5*t)/60
or
a=arc cot((H0+5*t)/60)

and

tan(a)=60/(H0+5*t)
or
a=arctan(60/(H0+5*t))

to get the rate of declination of the angle, take the first derivative:
d/dz of arctan(z)
=1/(1+z^2)

also, use the quotient rule


j

Answer 2

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 'θ'.

As you know: x² + y² = r²

Will come back to that.

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

Take the derivative with respect to time: dx/dt = (dr/dt)Cos(θ) - rSin(θ)(dθ/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(θ) - rSin(θ)(dθ/dt) (you want the rate of change of the angle with respect to time)
==> (dθ/dt) = [(dr/dt)Cos(θ) - dx/dt]/[rCos(θ)]

Plug your numbers in and you should see that:

(dθ/dt) ≈ 0.118°/s

«« done »»

This seems like a reasonable result to me. So look into my work and make sure I didn't make any mistakes. I looked it over, and I don't see anything that stands out, but that doesn't mean I didn't miss something obvious. If nothing else maybe it will get you going in the right direction. What a fun, fun problem. Probably one THE most interesting related rates problems I've seen in a long time. Hope this helps.

Answer 3

sorry