Question

Rate of change question!

Answer 1

PLease move to Q9 (iii)
http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s02_qp_1.pdf

Answer 2

Please "explain".
Only answers NOT required.

Answer 3

Mini?

Answer 4

the slope of the tangent line at point P is the derivative evaluated at x=1\[\frac{dy}{dx}|_{x=1}\]
what is that in this case?

Answer 5

i did the first and second part. want to understand the 3rd one.

Answer 6

ok so you need the equation you got from part 2
differentiate implicitly wrt t

Answer 7

http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s02_ms_1.pdf
Move to last page for the answers.

Answer 8

oh their solution is smarter than mine...
so what do you need?

Answer 9

How to do the third part?

Answer 10

Lol, the solution is already there..
Use the chain rule as shown, and sub it in.

Answer 11

i want to know how they substituted the values?

Answer 12

i did this in May 2011. im blank now.

Answer 13

you have m=dy/dx=4/3 from part 1
and they give you dx/dt in the problem

Answer 14

why they took 4/3 ?

Answer 15

that is\[frac{dy}{dx}|_{x=1}\]which is the slope of the tangent, m

Answer 16

\[\frac{dy}{dx}|_{x=1}\]

Answer 17

Because its the gradient dy/dx and you needed it to sub it in for the chain rule.

Answer 18

so.. ?

Answer 19

And what's x=1?

Answer 20

do you agree that\[\frac{dy}{dx}|_{x=1}=4/3\]?

Answer 21

Yes

Answer 22

We found that in the first part.

Answer 23

and do you agree that\[\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}\]?
(by the chain rule)

Answer 24

Yes..

Answer 25

and we are talking about the point x=1 according to the problem
hence, we use the value for dy/dx at x=1 (which is 4/3)
and dx/dt (which we are given in the problem)
to get\[\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=\frac43\cdot0.3=0.4\]

Answer 26

I think i got it. Thanks.
i will do similar problems, then i think i will polish on this

Answer 27

I'm sure it'll click
see ya!

Answer 28

Thankssss.