Hey guys
Could someone please explain to me how to do the questions attached? I've been having a hard time figuring out these ones. Sorry if it looks like it's a lot.
Thanxx a lot.
001a) We have an expression for I in terms of x.
I=[b^2+(x+8)^2]^-1+[b^2+(x-8)^2]^-1 -
(So i just made the denominator the numerator with indice laws)
We can now Differentiate this easily: (becareful though as it is a function of a function)
dI/dx=-[b^2+(x+8)^2]^-2 x 2(x+8) x (1) -1[b^2+(x-8)^2]^-2 x 2(x-8) x (1)
=-2(x+8)/[b^2+(x+8)^2]^2-2(x-8)/[b^2+(x-8)^2]^2
Find the LCM,
=-2(x+8)[b^2+(x-8)^2]^2-2(x-8)[b^2+(x+8)^2]^2/[b^2+(x-8)^2]^2 x [b^2+(x+8)^2]^2
=-2P/Q
ii) Just Find Where the Max T.P is, show that it's a max at that point etc
If you have anymore trouble with it let us know
for b) Hint: Let Pm=y (or some other pronumeral), There is two expressions that can be made for XP, can you somehow find an over all expression for the length of the wire and substitute?
002ai)
V(cone)=(1/3)pir^2h
Since triangle is right angle we could look at using basic trig or pythagoras,
c^2=a^2+b^2
a^2= r^2+ (x-a)^2 [the line to the top of the cone and the "a" line are equal as they are both radii]
a^2=r^2+x^2-2ax+a^2
r^2=a^2-x^2+2ax-a^2
therefore r^2=(-x^2+2ax)
And as we know height= x (given)
So by substitution into the volume formula:
V=(1/3)pi(2ax-x^2)x
=(1/3)pi(2ax^2-x^3), As required
ii) Just differentiate and show its a max as normal
002b) I'll leave this for someone else but if noone else does it i'll do it tomoz or something
