Question

A curve is such athat {dy/dx}= {12 over (2x+1)^2}
and P(1,5) is a point on the curve.

(i) The normal to the curve at P crosses the x-axis at Q. Find the coordiantes of Q.

Answer 1

slope of normal = -1/ (dy/dx)
you've got the slope and point so find he equation for the normal and you can find the coordinates of Q

Answer 2

Hmm.. normal is perpendicular to tangent. dy/dx is the slope of tangent. Do you understand what i've written?

Answer 3

the slope of the normal meeting the curve = -1
so
m1m2= -1

-1/dxdy= m2 (the slope of the normal)

Does this equal to

(2x+1)^2/12?

Answer 4

you've forgotten the negative sign..

Answer 5

Oh, yes. - (2x+1)^2 /12

Answer 6

Then I have to find the equation of the line? How?

Answer 7

point - slope form

Answer 8

y=0 no?

Answer 9

(y-y1) = m(x-x1)
(x1,y1) = (1,5)

Answer 10

m = - (2x+1)^2 /12
and put x=1into m=...

Answer 11

Hm, Ok...

m= -(2x+1)^2/12
m= -9/12
m= -3/4?

Answer 12

should be like that

Answer 13

So, then

y=-3/4x+b
Then do I plug in the y and x values of (1,5)?

Answer 14

(y-y1) = m(x-x1)
(x1,y1) = (1,5)
Sorry if i've used the wrong words...

Answer 15

I mean, would the equation of the line be

y= -3/4x+5 3/4?

Answer 16

shouldn't it be y= -3/4x+4 3/4?

Answer 17

Why?

Answer 18

the m value is -3/4, so if you move that, it's 5+3/4

Answer 19

Sorry my mistakes :(

Answer 20

That's OK... So then do you plug in 0 in the y value?

Answer 21

to get x, when it's Q

Answer 22

yup!

Answer 23

Can you see what you get and so I can double check my answer?

Answer 24

I got 7.67 = x?

Answer 25

something ugly, 7 2/3, 7.67 if you need a decimal number

Answer 26

Yes. It looks ugly... I hope it's correct! :)

Answer 27

The next question says find the equation of the curve?....?

Answer 28

use integration

Answer 29

\[y=\int\limits dy/dx\]

Answer 30

I thought we have the slope of the curve, which is 12/(2x+1)^2?
So, can't we just use that? Why do we have to integrate?

Answer 31

Ehhh.. you need the curve right?

Answer 32

I need the equation of the curve...

Answer 33

then you need to know that dy/dx is the slope of the curve, to find the curve you need to do the reverse..

Answer 34

Oh... I just thought you could use that slope and do
y=mx+b, where m is the slope??

Answer 35

nope... it's a curve, not a straight line. For a curve, the slope differs from point to point. So, there are no constant slope

Answer 36

*there is

Answer 37

Oh, Ok. So.... how do you integrate it?

Answer 38

I'm not good at using math latex here.. give me some time to write it..

Answer 39

Sure. I'll be back in a moment.

Answer 40

am back

Answer 41

Thank you! So, is that the equation?

Answer 42

Wait, why is it timesed by 1/2?

Answer 43

Personally, i think it is, but not sure

Answer 44

Should I ask Sam?

Answer 45

you can d/dx(equation) and see if you understand why it is multiplied by 1/2

Answer 46

Sure,you should , I though he should have taught you since you asked the question.. but that falls short of my expectation :(

Answer 47

@.Sam. Can you check all these answers.?

Answer 48

Callisto, there's a last part of the question.

Answer 49

Honestly, i've learnt differentiation for a year and integration for half a year.. So , i'm not really good at that :S (no time to practise)

Answer 50

Um... the last part of the question says, A point is moving along the curve in such a way that the x=coordinate is increasing at a constant rate of 0.3 units per second. Find the rate of increase of the y-coordinate when x=1

Answer 51

your integretion was correct :)

Answer 52

Um, I have to go very soon. Can you see if you can help me with the last question, and then it'll be good?

Answer 53

Sorry was away from computer.
I'm not sure for this problem
from the question, you'll have dx/dt = 0.03
and you can do it like that
dy/dt = (dy/dx)(dx/dt) to find dy/dt..

Answer 54

sorry again, it should be dx/dt = 0.3

Answer 55

then it's like
dy/dt = {12 over (2x+1)^2} * 0.3
Put x=1into dy/dt and you can find the answer, i think