Parametrics

Here is the question which I was stuck on right from the beginning. And here is the answer which I can’t understand

It’s the first line actually. Why is t = pi/6. If it were pi/6 then subbing it into y= a sint ( diagram indicates that y = 1/2) then that would make a = 1 but later on in the answer it works out that a = 6.796

I’m really stuck with tis and any clue would be appreciated.

Thanks

Youd have to post the rest of the solution, but it looks like the diagram should be +/-a/2 on the y axis, as (0,a/2) is the point they use in the equation of the tangent.

This is the best I could do for getting all the solution in one photo


thanks

As in the previous post, it looks like a typo on the graph and the y axis should be +/-a/2.

As in the previous post, it looks like a typo on the graph and the y axis should be +/-a/2.

Thanks

I've been working on that question but I'm stuck again now. When the integration to find the areas are concerned why does the integral for the curve have an upper of pi/6? At B y =0 so asint is 0 so t = 0, pi, 2pi etc. So as x = a cos3t. I have no idea where pi/6 comes from .

The upper part (symmetric) of the curve is from t=0 (cuts x-axis) to t=pi/6 (cuts y-axis), so the shaded area is
area under tangent - area under curve
with the latter expressed parametrically in terms of t.

HI

I get where the area is found from. In my working I didn't integrate the tangent I just worked out the are of the triangle 0.5(bxh). It's the area under the curve that I I'm having a tiny bit of trouble with . I know I'll have to integrate ydx and need to get dx in terms of dt and replace y by asint but its' the upper limit of integration that foxes me. The curve crosses the x axin when y = 0. y=a sint so t = 0, pi...... sub those values into x = a cos 3t and I don't get pi/6 for the upper limit. That's the only bit now that's causing a problem.

Could you please explain the upper limit value. Thanks

area of tangent or area of triangle is obv. the same thing. For a parametric curve
Int y dx = Int y dx/dt dt
so its basically the chain rule as y and x are functions of t. So
https://www.savemyexams.co.uk/a-level/maths_pure/edexcel/18/revision-notes/9-parametric-equations/9-2-further-parametric-equations/9-2-2-parametric-integration/
Obv, the limits in the first one are in terms of x and in the second one are in terms of t. t=0,pi/6 correspond to the axes crossing points as youd want or acos(3t)=0,a in terms of x.

You could get y(x) and do the first integral, but the second looks more direct here.

I have understood everything you have said but I still can't see why the upper limit of integration of the curve is pi/6.

I have understood everything you have said but I still can't see why the upper limit of integration of the curve is pi/6.

You may find that t = pi/6 is the lower limit (0, a/2) and t = 0 is the upper limit (a, 0).

Thank you. All is now explained. I can’t remember having to do such a thing before. Just hope that I will remember in the future. It does seem odd to have the lower limit larger than the upper so perhaps that will be the trigger to my memory.

In x-y terms, you're integrating from "left to right" to find an area (smaller x value is the lower limit, larger x value is the upper limit). It may look counter-intuitive at first, but if you're working in parametric coordinates it's quite possible for the larger t-value to correspond to the leftmost x coordinate.

I have done so many parametric integrations over the years (still plodding along) I might have come across it before but, as you say, it looks counterintuitive.