problem (1 Viewer)

[gronkboyslim]

sketch the graph y^2 = x^2(1-x) and find the area of the loop


i keep getting 4/15
but the answer is 8/15

but i also sqrted both sides

and worked with y = x sqrt(1-x) which i dont know if im allowed

 

[kini mini]

Originally posted by gronkboyslim
sketch the graph y^2 = x^2(1-x) and find the area of the loop


i keep getting 4/15
but the answer is 8/15

but i also sqrted both sides

and worked with y = x sqrt(1-x) which i dont know if im allowed

The loop? Any boundary values, or are you describing a diagram? I presume you mean the part of the function from (0,0) to (1,0)??
(thinking as I type hehe)

Let's have a look at this curve. What do we know? Since it is y^2=..., it will *always* be greater than or equal to 0. This is also obvious from the range I put above. So the reason square rooting works is that you do not have to consider a negative RHS.

 

[spice girl]

did you double the answer cos you need to consider the area of the whole loop, not just the half above the x axis?

 

[Lazarus]

Originally posted by gronkboyslim
and worked with y = x sqrt(1-x) which i dont know if im allowed

That's completely fine - just remember to take both positive and negative square roots. You need to graph y = -x sqrt(1-x) on the same graph, which is in fact the same curve reflected over the x-axis.

The loop is between x = 0 and x = 1. To find the area, it's probably best to integrate just the positive version of the curve between 0 and 1 and double the result (which is why the answer is 8/15 and not 4/15).


edit: gah, spice girl beat me. well done.

 

[gronkboyslim]

ahh i see thanx

but i thought the sqrt of x^2 was absolute x?

 

[kini mini]

LOL I totally forgot about that, brain is obviously leaking out very fast post HSC!