here is the problem:

A particle moves along y=x^3 in the 1st quadrant in such a way that its x-coordinate increase at a steady 12 m/s. How fast is the angle of inclination (theta) at the line joining the particle to the origin when x=2?

 

here is what I currently have

dx/dt=12

dy/dt=2x^2 * dx/dt

x=2

y=8

 

I can't find a relation to solve for derivative of theta.

For this problem, you want to keep in mind a few things. One, that y=x^3 which means that the triangle made from the two lengths can be put into terms of x. B) the tangent of the angle is y/x. Fourthly, I'm assuming that the axes are measured in meters. Many problems require a change of units before the actual calculations. With that in mind, we can solve this problem.

 

tan (angle) = y/x

y = x^3

 

So tan (angle) = x^2

From here, we'll take the inverse tangent of both sides to make the equation in terms of the angle.

angle = inverse tangent (x^2)

 

From here, we'll take the derivative and see what happens.

d angle / d time = (1/(x^4 + 1)) * 2x * d x/ d t

 

This is really messy so let me explain. The derivative of inverse tangent of x is 1/(x^2 + 1). Using the chain rule twice, we get the above.

 

At this point we'll plug numbers in and find d angle / d time. We know x, as it is given as 2. We know d x/ d t because it says the particle's moving at 12 m/s.

 

I get 48/17 radians per second.

how do you use chain rule twice to find the derivative of the inverse tan(x^2)?

Sorry, my wording is confusing.

First you'll need to know what the chain rule is. You might've been taught it a different way than me, although the general concept will be the same. Derivative of the outside times the derivative of the inside was how I was taught it. It seems like you're familiar with it. Secondly, you'll need to know how to differentiate inverse tan(x).

The derivative of that is [math]\frac{1}{x^2 +1}[/math]

 

Normally, if a problem required you to differentiate inverse tan(x^2), you'd use the chain rule once.

So it'd be [math]\frac{1}{\left(x^2 \right)^2 +1} * 2x[/math]

 

Hehe, let me try out latex.

 

Anyways, you would differentiate inverse tan(x^2) that way only if x was an independent variable. However, this is a related rate problem and it says that x is dependent on time. That means we'll have to differentiate x, which is [math]\frac{dx}{dt}[/math] and multiply it to the derivative up there (chain rule).

 

So in the end, what you end up with is

[math]\frac{1}{\left(x^2 \right)^2 +1} * 2x * \frac{dx}{dt}[/math]