Just to get you going. For 1), you know that , so (somewhat heuristically) we get
and remembering ,
.
Integrate for to get for some , and use the given initial condition when to get .
Hi
I'm having some trouble working out how to convert from intrinsic to cartesian coordinates, and from cartesian to intrinsic, when the function is not given explicitly in terms of y. Let me show you a couple of problems which I have been unable to solve to illustrate this,
(1) The intrinsic equation of the curve is . The curve passes through the point (2,0). Show that and find y in terms of
(2) For the curve with equation , where s is measured from the point , show that
For the first I tried to let and then take the cosine of both sides since however I ended up with a bit of a mess and didn't really get anywhere.
For the second I tried to find an expression for and set it equal to and also find the length of s, but I ended up with a horrible integral and again didn't really get anywhere.
I'm really at a loose end on how to solve these now, I think there must be something simple I'm missing, but I just can't see it.
Any help would be much appreciated, thanks
Stonehambey
A little less heuristically:
(And first some clarification. here is the angle the line from (0,0) to a given point on the curve makes with the positive x-axis (so is the slope of the tangent line- the derivative of y with respect to x) and s is the arc length to that point from some fixed point on the graph.)
so
since . But so we have the differential equation
so
Integrating the right side by parts, using , , we have [tex]du= 2d\psi[tex], [tex]v= sin(\psi) so
.
Using the fact that x= 2 when (the point (2, 0) is on the x-axis, of course, so the line from (0,0) to (2,0) makes angle with the x-axis) C= 0 giving as required.
By the chain rule, and we already know that so and so we have
and you can again use integration by parts to find y as a function of .