C2 Help: Differentiation

Please help me with the following questions:


1.The fixed point A has coordinated (8, -6, 5) and the variable point P has coordinates (t, t, 2t).

(a) Show that AP² = 6t² - 24t +125

(b) Hence find the value of t for which the distance AP is least.

(c) Determine this least distance

2. A cylindrical biscuit tin has a close-fitting lid which overlaps the tin by 1 cm. The radii of the tin and the lid are both x cm. the tin and the lid are made from a thin sheet of metal of area 80pi cm square and there is no wastage. The volume of the tin is V cm³.

(a) Show that V=pi (40x - x² - x³)

For some reason I got V=pi (40x - x²)

Besides the two above questions, there are some past threads (just a few threads away) that needs replies from math geniuses.

(a)The square of the distance between the points $(a,b,c)$ and $(d,e,f)$ is $(a-d)^{2}+(b-e)^{2}+(c-f)^{2}$
(b) Complete the square to give $AP^{2} = (t-g)^{2}+h$ and obviously then the value of t is g when the distance is the least as $(t-g)^{2} \geq 0 \forall x \in R$
(c) Using the value of t acquired in (b) we have $AP^{2}=h$ and so $AP=\sqrt{h}$

Thanks for telling me how to prove for (a)

but for (b) I used differentiation, in that i found f'(AP²) and solved f'(AP²) = 0.

That works fine though I assume you meant you found (AP²)' and solved (AP²)'=0 for t? In any case the difference is purely notational.
It's a bit like using a hammer to open a cereal packet though

Top and bottom 2?x²
overlap 2?x
side 2?xh

Total: 2?(x²+x+xh) = 80?
Get h = (40-x²-x)/x
V = ?x²h = ?(40x-x³-x²)

Aitch

huh? you mean you don't use f'(AP) but use (AP)' explain!!!! What's the diff?

I'm a bit slow... can you pls elaborate!

Well you have have AP² as a function of t but you haven't defined what 'f' is and so your notation is senseless.
I assume though that you defined AP² as f(t).
We then have (AP²)'=f'(t)
You aren't finding f'(AP) though, you're finding the values of t for which f'(t)=0.

True. Thanks for correcting me. So I guess f'(t) = (AP)' = dAP²/dt

You need to eliminate h

You work out how much metal sheet you need for
the discs top and bottom 2?x²
the overlap of the lid 2?x (*1)
the side 2?xh

total sheet metal needed = 2?(x²+x+xh)

You know this is 80? (cm²)

so 2?(x²+x+xh) = 80?

You now have to get rid of h (it's always like this in these questions!)

rearrange to get:h = (40-x²-x)/x

The volume is then ?x²h = ?(40x-x³-x²)

Aitch

That's it

It's rather annoying that the minimum message length is 15 characters (that in itself is the purpose of this rather useless line).

OMG: I forgot what the discs top and bottom and the overlap completely!! I'm so stupid! Thank you for all your effort. Rep for you.

No problem.

This process of eliminating h (or another variable) is very typical of these questions.

MY favourite mistake was in the next bit, when you get dV/dx and set it =0 to get the max volume.

...Several times I've put "So radius of x= (10/3) gives the maximum volume" as my answer when it is the maximum volume itself which is asked for!

Aitch