The Student Room Group
Reply 1
Pythagoras comes to the rescue - in that OP2=x2+y2 \mathrm {OP}^2 = x^2 + y^2 - you already have an expression for yy... there's no differentiation necessary.
Reply 2
is that some sort of rule, haven't learnt it or is it something really obvious which I don't see :smile: lol
Reply 3
Well - you have the formula y=512x2 \displaystyle y = 5 - \frac{1}{2}x^2 . So, if you go x x along the x-axis, you will have to go 512x2 5 - \frac{1}{2}x^2 along the y-axis to meet the curve.

You have a point, P, which is at coordinates (x,y). What we need to know is the distance from the origin (0,0) of this point. We end up with a right-angled triangle: one side will be x x , and one side will be y y (which is the same as 512x25 - \frac{1}{2}x^2 ). The hypotenuse (the longest side of the triangle) will be the distance from the origin to the point P.

Pythagoras' Theorem is that 'the square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the other two sides' - this means that the squares of the lengths of the two shorter sides of the triangle equals the square of the longest side. Put into an equation, this is x2+y2=z2 x^2 + y^2 = z^2 .

Your question is asking you to find z2z^2 (OP2\mathrm{OP}^2 is the square of the hypotenuse of our right-angled triangle) but it's no use just saying that it's equal to x2+y2 x^2 + y^2 ... you have to use the expression for y in terms of x. That is to say... find x2+(512x2)2 x^2 + (5 - \frac{1}{2}x^2)^2 .

Any clearer? (And more importantly, do you understand what we're doing?)
Reply 4
Yes definitely much more clearer and I kind of get what were doing, differentation is my weakest topic on C2 thanks a lot :smile:
Reply 5
hi iam on the same excerise iam just wondering has any 1 done question 5? if so i need help !! please
Reply 6
bigbosss
hi iam on the same excerise iam just wondering has any 1 done question 5? if so i need help !! please

For question 5:

a) Rearrange the perimeter equation in terms of y; note there’s only one x side.
b) Substitute the equation in y from part a) into the equation you need to find the area. This gives you the area in terms of x which is what you really want.
c) Area in given in squares, it is a quadratic. This means there is only one maximum. At the maximum the gradient is always 0.

Is that any help? :]
Reply 7
Original post by Guinny
For question 5:

a) Rearrange the perimeter equation in terms of y; note there’s only one x side.
b) Substitute the equation in y from part a) into the equation you need to find the area. This gives you the area in terms of x which is what you really want.
c) Area in given in squares, it is a quadratic. This means there is only one maximum. At the maximum the gradient is always 0.

Is that any help? :]


I have been stuck on part b) for ages now!!! the formula for the area of a circle = π × r^2 , because we have x but no r, should i just write π(1/2x)^2 ? How would i rearrange that? Please help! Thaaanks :smile:
Original post by Rusnytele
I have been stuck on part b) for ages now!!! the formula for the area of a circle = π × r^2 , because we have x but no r, should i just write π(1/2x)^2 ? How would i rearrange that? Please help! Thaaanks :smile:


I don't know your question. But if your simplifiying:

Pi (1/2x)^2 = Pi(1/4x^2)

= Pi/4x^2
Reply 9
does anyone have the solution to question 1 and 3? i can't work them out!