caseclosed Posted November 20, 2005 Share Posted November 20, 2005 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. Link to comment Share on other sites More sharing options...
psi20 Posted November 20, 2005 Share Posted November 20, 2005 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. Link to comment Share on other sites More sharing options...
caseclosed Posted November 20, 2005 Author Share Posted November 20, 2005 how do you use chain rule twice to find the derivative of the inverse tan(x^2)? Link to comment Share on other sites More sharing options...
psi20 Posted November 20, 2005 Share Posted November 20, 2005 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] Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now