Yes there is a shorter solution. Consider this diagram:

Do you agree that triangles OAR and OAP share a common height h₁?

Similarly triangles OAR and OAQ share a common height h₂.

From your usual triangular area formula, the only thing that is different are the bases (lying along OP and AQ respectively), but we can express this using the ratio values h and k obtained in part (4). Hence,

Using the value of obtained in part (4),

Using the value of obtained in part (4),

=

Short enough for you?

Remember what was said in this blog entry about vector questions often asking for something a little bit off-tangent?

Your workings for part 5 demonstrates this point perfectly. You started off well (i.e. you were looking for common heights and all that) but then it all became tedious once you start substituting in the vectors. Imagine this in exam conditions. Many students tend to get sucked so deep into the vector 'pit' (afterall you've been solving for vectors till this point) that they've forgotten that sometimes a different approach is called for.

Hope you've learned something from this today. I know you're excited but please, please refrain from swearing here again!

Miss Loi: Now, now ... before you get overly hysterical let's keep this blog a family-oriented place shall we?

(area of OAP) / (area of OAQ)
= (4/3 x 1/2 x OR x AR) / (1/2 x OR x AQ)
after canceling

= (4/3 x AR) / (AQ)
(AR = kAQ)
= [4/3(3kb/7 - ak)] / (3b/7 - a)
sub k as 7/16
= (12b - 28a) / 48) x [7 / (3b -7)]
point to note 12b - 28a is 4 times of 3b-7a

hence [(12b - 28a)28] / [(12b - 28a)48]
after cancelin..
28/48 = 7/12

(area of OAP) / (area of OAQ) = 7/12

forbearance is appreciated if i got it wrong-.-

3/7k = 1/4h (2)
sub (1) into (2)
though k isnt needed but i find it to be useful in part 5 so>.>

Miss Loi: Congratulations! You've got part (4) done nicely - cleaned up your workings a little for better readability! But why are the workings for part (5) spanning across 2 comments??? ... hmmmm ... let me see ...

area of OAR / area of OAP
let h be AR (both share the same lenth for height)

= (!/2 x OR x h) / (1/2 x OP x h)
after cancelation
= OR/OP
= sub in OR and OP from earlier sections and sub in k as 7/16
= (1a - ka 3/7kb) / ([3a b] / 4)
= (9a 3b)/16 x 4/(3a b)
notice 3a b is basically 1/3 of 9a 3b so after combin and canceling we'll get 3/4
area of OAR / area of OAP =3/4(god this question is tedious there has to be a simpler method)

So now you have two equations for V:

(from part i)

So when V = 36 cm³, simply sub in this value to find the corresponding h and t:

cm

Factorizing this quadratic equation,
(t + 8)(t - 24) = 0
You'll get:
t = -8 (NA), 24
∴ t = 24 s

Now the question asked for the rate of change of the h i.e. when the V = 36 cm³.

Hence using the Chain Rule (from your A-Maths Rate of Change chapter):

Now you can easily differentiate from the given equation, while you've already obtained from part (i), so:

Sub in the values of h and t you obtained earlier for V = 36 cm³:

Using your super-duper scientific calculator to calculate the above, you should be able to get

Wow that was some typing! To reward Miss Loi the student from answering your super-duper cheem question, can Teacher Kiroii do parts 4-5 of the original vector question for her?

P.S. This is basically an A-Maths question being answered in an E-Maths blog entry. To prevent potential confusion to readers, try asking A-Maths questions in one of Miss Loi's A-Maths blog entry ok?