Well I'll post the solutions to Q10 now (I like working backwards
). Watch this space for like 3 mins
Question 10.
a) i)
log<sub>e</sub> e<sup>2ax</sup>
=2ax*log<sub>e</sub> e (log<sub>b</sub> c<sup>d</sup>=d*log<sub>b</sub> c)
=2ax(1) (log<sub>a</sub> a=1)
=2ax
ii)
<sup>a</sup>∫<sub>0</sub> log<sub>e</sub> e<sup>2ax</sup> dx
=<sup>a</sup>∫<sub>0</sub> 2ax dx (from i))
=[2ax<sup>2</sup>/2]<sup>a</sup><sub>0</sub>
=[ax<sup>2</sup>]<sup>a</sup><sub>0</sub>
=a<sup>3</sup>-a(0)
=a<sup>3</sup>
b) i)
D=√((x<sub>1</sub>-x<sub>2</sub>)<sup>2</sup>+(y<sub>1</sub>-y<sub>2</sub>)<sup>2</sup>)
D<sup>2</sup>=(x-1)<sup>2</sup>+(y-4)<sup>2</sup> --- (1)
But y<sup>2</sup>=2x
∴ x=y<sup>2</sup>/2
Subbing x=y<sup>2</sup>/2 in (1)
D<sup>2</sup>=(1/2*y<sup>2</sup>-1)<sup>2</sup>+(y-4)<sup>2</sup>
As required.
ii) Now D is a distance and is always positive, ∴ the point of minimum of D<sup>2</sup> is also the point of minimum of D
dD<sup>2</sup>/dy=2(1/2*y<sup>2</sup>-1)y+2(y-4)
=2y(1/2*y<sup>2</sup>-1)+2y-8
=y<sup>3</sup>-2y+2y-8
=y<sup>3</sup>-8
=(y-2)(y<sup>2</sup>+2y+4) (Difference of two cubes)
For stat dD<sup>2</sup>/dy=0
∴
(y-2)(y<sup>2</sup>+2y+4)=0 for stat
y=2 for stat
check concavity
d<sup>2</sup>D<sup>2</sup>/dy<sup>2</sup>=3y<sup>2</sup>
at y=2, d<sup>2</sup>D<sup>2</sup>/dy<sup>2</sup>=12>0 ∴ it is a minimum.
Hence the minimum distance occurs at y=2.
iii)
Minimum distance occurs at y=2
Sub y=2 into D<sup>2</sup>=(1/2*y<sup>2</sup>-1)<sup>2</sup>+(y-4)<sup>2</sup>
D<sup>2</sup>=(1/2*4-1)<sup>2</sup>+(2-4)<sup>2</sup>
=1+4
=5
∴ D=√5 units (distance must be >0)
As required.
