Edexcel C4 - Integration of Parametrically Given Curves

I'm not really sure how to do these at all. Can someone please show me how?

The curve C has two arcs and the equations x = 3t^2, y = 2t^3 where t is a parameter.

Find an equation of the tangent to C at the point P where t = 2.

Worked out as y - 16 = 2(x-12)

The tangent meets the curve again at the point Q.

b) Show that the coordinates of Q are (3,-2)

Done.

The shaded region R is bounded by two arcs OP and OQ of the curve C, and the line PQ.

c) Find the area of R.

It seems like they want me to integrate between 0 and P or something, but the curve goes below the x axis so I'm not sure how.

Another question:

A curve has parametric equations x = 5cosa, y = 4sina where 0<=a<2pi.

Long story short;

At P, a = pi/4. And the equation at the tangent at P: y = -0.8x + 4root2

R is the point where the tangent meets the x axis. R is therefore (5root2,0).

The shaded region is bounded by the tangent PR, the curve and the x-axis. Find the area of the shaded region, leaving your answers in terms of pi..

Again, no idea how to do this.

For anyone with access to the Heinemann Edexcel C4 book, the first one is question 2 and the second one is question 15 from the exercise 6L on integration.

Reply 1

Kolya

I was asked about your first question by somebody in my school who is taking C4 on Monday; he could not get the answer in the back, and neither could I, so it may be incorrect. I shall show what I have done and if anybody disagrees then please shout.

At the point P, t = 2.
At the point Q, t = -1 (by putting the y value of Q into C, ie. -2 = 2t^3)

So

\displaystyle \int^2_{-1} y \frac{dx}{dt} \ dt = \int^2_{-1} (2t^3)(6t) \ dt = 12 \int^2_{-1} t^4 \ dt

\displaystyle = 12[\frac{t^5}{5}]^2_{-1} = 12[(\frac{32}{5})-(-\frac{1}{5})] = (12)(\frac{33}{5}) = \frac{396}{5} = 79.2u^2

OP: Hopefully you can use this to help you solve the second question by yourself?

Reply 2

Swayum

But is it right to integrate between those values because if you look at the graph it goes below the x axis so surely the integral is losing some area?

Also, for the second one, the tangent goes past the curve. I'm not really sure what the limits of my integral should be because the curve isn't defined where the tangent meets the x axis. Also, am I meant to do the integral of the area under the tangent minus the area under the curve?

Reply 3

Kolya

How is it losing some area? The negatives cancel so you are adding two postive terms in the limits.

Reply 4

Swayum

What do you mean the negatives cancel? Consider a y = x^3 curve. If you integrate from -5 to 5, wouldn't that give 0? Which shows that the area under the x axis is negative. So doing you integral, where some of the curve is below the x axis, should give a smaller area, shouldn't it? Could you also take a look at my other thread which is about the same question as the second question in this post?

Reply 5

Kolya

There is no problem with this question as your negative area is given as a negative "area" when limits are applied, and hence becomes an area added when we tidy up. Please look at what happens when the limits are applied and tell me where your supposedly smaller area is coming from? Integrate (-1,0) and (0,2) separately if you really want, but it will not make a difference.

Reply 6

silent ninja

I dont have a C4 book but dug out my working. There should be a picture of a graph for this?
Anyway, I get area= 16.2 for the first one, after taking into account area of the two triangles.

Reply 7

Kolya

silent ninja

I dont have a C4 book but dug out my working. There should be a picture of a graph for this?
Anyway, I get area= 16.2 for the first one, after taking into account area of the two triangles.

OOps. Sorry, you are correct.

Reply 8

Cataclysm

Excuse me but could you show me the last step.

How did you get from 79.2 to 16.2 ---> which two triangles are we talking about, I see one but I can't figure it out properly.

Reply 9

silent ninja

This was a year ago, i don't have the foggiest idea. If you could post up the image of the graph or something it might help. Or perhaps someone who has the book might be able to help!

Here is the image:

I integrated parametrically between 2 and -1 and got 79.2 and can see there is a triangle from Q to P but can't quite get the right answer.

Reply 12

silent ninja

Notice line PQ is not horizontal so when you integrate you get the area vertically down from P and to the left; this area up to the tangent/line is not wanted. It makes a triangle so work out it's area by figuring out where the tangent crosses x-axis.

Second, integrating between OQ gives you the area vertically to the left of OQ but if you notice once again, you miss out on a small triange. So draw a perpendicular line down to Q from the x-axis and figure out the area of this triangle and ADD it to your total.