The Student Room Group

Direct Proportion

x is directly proportional to the square of y.
y is directly proportional to the cube of z.
z = 2 when x = 32
Find a formula for x in terms of z

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x=ky^2
y=kz^3
32=k1 (k2* 2^3)^2
32=64k1k2^2
32/64 =k1k2^2
x= 1/2z^3
(edited 1 year ago)
Original post by esam 01
x=ky^2

EDIT - it is against forum rules to post a solution ...
Reply 3
I don't agree with esam 01's answer but I could be wrong. What do you have so far @sleepycatash?
do you have an answer
Original post by cmbh951
I don't agree with esam 01's answer but I could be wrong. What do you have so far @sleepycatash?

The mark scheme says its x=1/2 z^6 but i don't know how to get there
Original post by sleepycatash
The mark scheme says its x=1/2 z^6 but i don't know how to get there

i ended up with the same answer but z power 3 not 6
Reply 7
Original post by sleepycatash
The mark scheme says its x=1/2 z^6 but i don't know how to get there

A hint is to think about what directly proportional means. Can you write x and y in an equation together if you know that x is directly proportional to the square of y?
how is it z^6
Original post by sleepycatash
x is directly proportional to the square of y.
y is directly proportional to the cube of z.
z = 2 when x = 32
Find a formula for x in terms of z

Post your working and ignore what others have posted
Original post by esam 01
this is not the solution

You've tried to work through the problem - that's against the rules EDIT please
Original post by cmbh951
A hint is to think about what directly proportional means. Can you write x and y in an equation together if you know that x is directly proportional to the square of y?

x=ky^2
y=cz^3
x=k(cz^3)^2
x=kc^2z^6
32=kc^2 2^6
32=64kc^2
1/2= kc^2
(edited 1 year ago)
z is cubed
Original post by Muttley79
Post your working and ignore what others have posted

I hope I'm not included in "others" when I'm giving a genuine hint and not the full (incorrect) solution.

Original post by sleepycatash
x=ky^2
y=cz^2


Looks good, but y is directly proportional to the cube of z, not the square.

Now if we know what z is when we know what x is, we would want an equation involving only x and z.
Is my edit the right working? Am I on the right track?
yes
Original post by sleepycatash
Is my edit the right working? Am I on the right track?

Yep, you've got it. So going back to your formula for x in terms of z, is there a nicer way to write it?
Original post by cmbh951
Yep, you've got it. So going back to your formula for x in terms of z, is there a nicer way to write it?

Would it be x=1/2 z^6
Reply 18
Original post by sleepycatash
Would it be x=1/2 z^6

yes
most excellent
Original post by sleepycatash
Would it be x=1/2 z^6

That's it :smile: a nice check is to put your values of x and z back into the formula and see if it works.

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