A Kite 70 feet above the ground moves horizontally at a speed of 4 ft/s. At what rate is the angle between the string and the ground decreasing when 150 feet of string is out?

Question

johnweldon1993 · Answer

Well diagram it out first

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johnweldon1993 · Answer

You know it is moving horizontally at 4ft/s so $\large \frac{dx}{dt} = 4$

johnweldon1993 · Answer

Any thoughts from here? or still stuck?

anonymous · Answer

everything you just said is as far as I get

johnweldon1993 · Answer

Okay, so what we want next, is some expression for 'x' in terms of theta

Why? Well remember we eventually want to solve for $\large \frac{d\theta}{dt}$

So hmm...I see a relation $\large tan(\theta) = \frac{70}{x}$ right?

johnweldon1993 · Answer

If we solve that for 'x'...we have

$$\large x = \frac{70}{tan(\theta)}$$ right?

anonymous · Answer

right

johnweldon1993 · Answer

Okay good, so lets take the derivative of that wrt time

Oh we cant just yet...we need to use the chain rule

$$\large \frac{dx}{dt} = \frac{dx}{d\theta} \frac{d\theta}{dt}$$

Now we can

anonymous · Answer

where did that equation come from?

johnweldon1993 · Answer

That's just me rewriting something

Notice

$$\large \frac{dx}{dt} = \frac{dx}{\cancel{d\theta}}\frac{\cancel{d\theta}}{dt} = \frac{dx}{dt}$$

Not changing anything...but at the SAME time...it is allowing us to execute our derivative

anonymous · Answer

oh ok

johnweldon1993 · Answer

So now, what is the derivative with respect to theta?

$$\large \frac{dx}{d\theta} \frac{70}{tan(\theta)} = ?$$

johnweldon1993 · Answer

Sorry I actually gotta get going so I'm gonna do this out quick, I'll check back in a little bit to see how you did or if any questions arose

So Recap:

$$\large \frac{dx}{dt} = 4$$
Now take derivative of
$$\large x = \frac{70}{tan(\theta)}$$
With respect to theta

$$\large \frac{dx}{d\theta}\frac{70}{tan(\theta)} = -70csc^2(\theta)$$

Now remember where I had that new way of writing the derivative?

$$\large \frac{dx}{dt} = \frac{dx}{d\theta} \frac{d\theta}{dt}$$
Plug in what we have

$$\large 4 = -70csc^2(\theta)\frac{d\theta}{dt}$$

Here, it is important to see at this moment we are solving for, the opposite of theta is 70 and the hypotenuse is 150...why is that important? Well

$$\large 4 = -70(\frac{1}{sin^2(\theta)})\frac{d\theta}{dt} $$

We know that at that exact moment...sin(theta) = 70/150 = 1/2

So that means sin^2(theta) = 1/4

So what we have now is

$$\large 4 = -70(\frac{1}{\frac{1}{4}})\frac{d\theta}{dt}$$

Which yes is

$$\large 4 = -70(4)\frac{d\theta}{dt}$$

Meaning

$$\large \frac{d\theta}{dt} = -\frac{4}{280} = -\frac{1}{70}$$

Which is some decimal which is your answer

*I'll check back in in a bit to see what you have :)

anonymous · Answer

it wasn't right