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P3 Differentiation watch

1. Could someone help me with the following two questions. I'm not sure how to go about either of them, although I know the general methods for solving questions like the 2nd one (ie. you use the chain rule), it's just I can't seem to work it out! Thanks

1) By using differentiation find approximate value for (27.005)^[2/3]

2) A variable rectangle is inscribed in a circle of diameter 20cm. At a certain instant one side is 16cm long and it is increasing at a constant rate of 0.75cm/s. Find the rate at which the shorter sides are changing.
2. (Original post by Hoofbeat)
Could someone help me with the following two questions. I'm not sure how to go about either of them, although I know the general methods for solving questions like the 2nd one (ie. you use the chain rule), it's just I can't seem to work it out! Thanks

1) By using differentiation find approximate value for (27.005)^[2/3]

2) A variable rectangle is inscribed in a circle of diameter 20cm. At a certain instant one side is 16cm long and it is increasing at a constant rate of 0.75cm/s. Find the rate at which the shorter sides are changing.
Is the answer to Q2 -1 cm/s? I'll post why if it's right!
3. (Original post by Bezza)
Is the answer to Q2 -1 cm/s? I'll post why if it's right!
YES! WELL DONE! Plese show me thanks! <makes note to give Bezza rep for answering so quickly>
4. Sorry I took so long - confused myself with letters when I tried to change them all for typing up! If you label the diagram as shown (a and s are the lengths of the whole sides), using right angled triangles you get a = 2*10sinx, s = 2*10cosx. Initial conditions are a = 16, da/dt = 3/4 cm/s. We need ds/dt = ds/dx*dx/dt

As a = 16, sinx = 4/5, cosx = 3/5, tanx = 4/3
ds/dx = -20sinx
At a = 16 cm, ds/dt = -20*4/5*dx/dt = -16*dx/dt

da/dx = 20cosx
dx/dt = dx/da*da/dt = 1/(20cosx)*3/4 = 3/(80cosx) = 3/(80*3/5) = 1/16 rad/s so ds/dt = -16*1/16 = -1 cm/s
Attached Images

5. For the first part, there is a specific approximation equation using a derivative ... can't remember it but you put f(x) equal to x^(2/3) and use some substitution to get it... I did that a while ago but couldn't find it on a search.
6. (Original post by Hoofbeat)
1) By using differentiation find approximate value for (27.005)^[2/3]
I've been thinking about this and although I can't remember I think this may be what they want
Let y = x^(2/3)
dy/dx = 2/3*x^(-1/3) approximately equal to change in y/change in x (delta y/delta x) for small x so delta y = 2/3*x^(-1/3)*(delta x) (approximately)
If you let x=27 and delta x = 0.005, (27.005)^(2/3) = y + delta y = 27^(2/3) + 2/3*27^(-1/3)*0.005 = 9 + 2/9*0.005 = 9.00111 (approx) which you can check on a calculator

Also, was it you who neg repped me for my earlier post? The comment said it was helpful and had a smily but I got a red gem and lost 3 points
7. (Original post by Bezza)
Also, was it you who neg repped me for my earlier post? The comment said it was helpful and had a smily but I got a red gem and lost 3 points
Thanks for your help! I repped you only about 15mins ago (couldn't do it earlier as had to wait 24hrs) and wrote "Thanks for all your help" and I'm sure I clicked "approve of post"!!! If it was I'm really really really sorry and I'll rep you every day for the next few days! lol
8. (Original post by Hoofbeat)
Thanks for your help! I repped you only about 15mins ago (couldn't do it earlier as had to wait 24hrs) and wrote "Thanks for all your help" and I'm sure I clicked "approve of post"!!! If it was I'm really really really sorry and I'll rep you every day for the next few days! lol
It was you then - at least now I can say I've had some neg rep! Don't worry about it though, it doesn't matter!
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9. (Original post by Bezza)
It was you then - at least now I can say I've had some neg rep! Don't worry about it though, it doesn't matter!
LOL!
Nice one Chloé!
I think you should take a break from all this revision! It's clearly making you worse!
10. (Original post by mik1a)
LOL!
Nice one Chloé!
I think you should take a break from all this revision! It's clearly making you worse!
YAY!!! You got my accent!!! It usually takes people years to work out how to do it!!!

I'm happy now. I think this revision is making me v.ill, so I saw the doctor and he prescribed me 5wks of bed-rest doing nothing but watching TV and playing on the computer (and riding my horse!). Think I'll have to cancel all my revision then! lol
11. (Original post by Hoofbeat)
YAY!!! You got my accent!!! It usually takes people years to work out how to do it!!!

I'm happy now. I think this revision is making me v.ill, so I saw the doctor and he prescribed me 5wks of bed-rest doing nothing but watching TV and playing on the computer (and riding my horse!). Think I'll have to cancel all my revision then! lol
Maybe the horse should have some bed-rest too

MB
12. (Original post by musicboy)
Maybe the horse should have some bed-rest too

MB
funny you should say that

he was on box rest for 2wks in may and only started goin out in field last fri for just an hour at time as he'd sprained sum muscles in his hock joint, so i can't ride him for another week at least. but he'll be better for the summer thnakfully :d
13. (Original post by Hoofbeat)
funny you should say that

he was on box rest for 2wks in may and only started goin out in field last fri for just an hour at time as he'd sprained sum muscles in his hock joint, so i can't ride him for another week at least. but he'll be better for the summer thnakfully :d
ahh, poor horsey, wish him well from me.

MB

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