Kyle Posted December 10, 2006 Share Posted December 10, 2006 [math]x=81-t^2[/math] [math]y=t^3-16t[/math] After asking about tangent lines, the question asks "The curve makes a loop which lies along the x-axis. What is the total area inside the loop?" I thought I was supposed to combine these equations so that there were no t's and then integrate it. I found that [math]y=(81-x)^{3/2}-16\sqrt{81-x}[/math] And, after looking at the curve, it looks like the answer for the Area should be [math]\int_{65}^{81} [(81-x)^{3/2}-16\sqrt{81-x}] \,dx=-4096/15[/math] I've tried this answer both negative and positive and it's been wrong. Maybe I haven't found the equation of the curve correctly or integrated correctly. Does anyone see my mistake? Link to comment Share on other sites More sharing options...
EvoN1020v Posted December 10, 2006 Share Posted December 10, 2006 Upon graphing those two functions together, I don't see any loop? Link to comment Share on other sites More sharing options...
insane_alien Posted December 10, 2006 Share Posted December 10, 2006 i also don't see a loop. Link to comment Share on other sites More sharing options...
the tree Posted December 10, 2006 Share Posted December 10, 2006 Upon graphing those two functions together, I don't see any loop?Isn't that one function, expressed parametrically? Anyways, I'll grab my textbook and have a proper answer for this in a bit... edit, got it: to find the area under a curve expressed parametrically you use [math]\int_a^b y \frac{dx}{dt} \cdot dt[/math] Link to comment Share on other sites More sharing options...
Kyle Posted December 10, 2006 Author Share Posted December 10, 2006 edit, got it: to find the area under a curve expressed parametrically you use [math]\int_a^b y \frac{dx}{dt} \cdot dt[/math] [math] \int_{65}^{81} [(81-x)^{3/2}-16\sqrt{81-x}] \,dx=-4096/15 [/math] [math] \int_{0}^{4} (t^3-16t)(-2t) \,dt=4096/15 [/math] I wonder if there might be a technical problem with this question. I think I have the right answer and your equation confirms it. I'll ask a few more people around the dorms and see if anyone has figured out what my teacher was trying to ask. 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