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Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Just a quick question (differentiation)

username1738583

\frac{\mathrm{d}y}{\mathrm{d}x} = 9 - x

a) Find the rate of change of

y

with respect to

x

when

x=3

b) Find the

x

coordinates of the two points where the gradient of the curve is 1

Can anyone help me with this? :redface:

:redface:

@Zacken or anyone else?

Scroll to see replies

Original post by homeland.lsw

\frac{\mathrm{d}y}{\mathrm{d}x} = 9 - x

a) Find the rate of change of

y

with respect to

x

when

x=3

b) Find the

x

coordinates of the two points where the gradient of the curve is 1

Can anyone help me with this? :redface:

:redface:

@Zacken or anyone else?

So, the rate of change at any given point

x

is given by

\frac{\mathrm{d}y}{\mathrm{d}x} = 9 - x

, the rate of change at

x=3

is given by plugging this into the above equation.

We notate this as:

\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} \bigg|_{x=3} = 9 - 3

username1738583

OP

Original post by Zacken

So, the rate of change at any given point

x

is given by

\frac{\mathrm{d}y}{\mathrm{d}x} = 9 - x

, the rate of change at

x=3

is given by plugging this into the above equation.

We notate this as:

\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} \bigg|_{x=3} = 9 - 3

So in simple terms

3 = 9 - x

???
therefor

x = 6

I'm not sure about this :sad:

:sad:

Original post by homeland.lsw

\frac{\mathrm{d}y}{\mathrm{d}x} = 9 - x

a) Find the rate of change of

y

with respect to

x

when

x=3

b) Find the

x

coordinates of the two points where the gradient of the curve is 1

Can anyone help me with this? :redface:

:redface:

@Zacken or anyone else?

That's part (a) dealt with, not part (b).

The gradient of the curve at a general point

x

is given by

\frac{\mathrm{d}y}{\mathrm{d}x} = 9 -x

, so the gradient is

1

precisely when

9 - x = 1

, solving this for

x

gives you the x co-ordinate.

Now we want the y-coordinate, but all we have is

\frac{dy}{dx} = 9-x

so what we do is integrate!

y = \int 9-x \, \mathrm{d}x

.

Can you take it from here?

Original post by homeland.lsw

So in simple terms

3 = 9 - x

???
therefor

x = 6

I'm not sure about this :sad:

:sad:

Nopes, you replace

x

with 3, and then

\frac{dy}{dx}big|_{x=3} = 9 - 3 = 6

. 6 is the gradient of the curve when

x=3

.

username1738583

OP

Original post by Zacken

Nopes, you replace

x

with 3, and then

\frac{dy}{dx}big|_{x=3} = 9 - 3 = 6

. 6 is the gradient of the curve when

x=3

.

The answer is -9 but why??

Original post by homeland.lsw

The answer is -9 but why??

How is the answer 9? :redface:

:redface:

username1738583

OP

Original post by Zacken

How is the answer 9? :redface:

:redface:

I dunno, I checked the answers to see if that gave me an idea and it said -9...

Original post by homeland.lsw

I dunno, I checked the answers to see if that gave me an idea and it said -9...

Have you copied the question down right? The correct answer is 6. :tongue:

:tongue:

username1738583

OP

Original post by Zacken

Have you copied the question down right? The correct answer is 6. :tongue:

:tongue:

yup dy/dx=9-x

find the rate of change of y with respect to x when x=3...

maybe a misprint?

Original post by homeland.lsw

yup dy/dx=9-x

find the rate of change of y with respect to x when x=3...

maybe a misprint?

Probably just the answers being wrong as usual, anyway. When x=3, dy/dx is 6. So that's part (a).

Original post by homeland.lsw

yup dy/dx=9-x

find the rate of change of y with respect to x when x=3...

maybe a misprint?

Are you sure it is a 3 and not 0?

Original post by zetamcfc

Are you sure it is a 3 and not 0?

Even then, the answers say -9

Original post by Zacken

Even then, the answers say -9

lol,such an idiot ! am, must be a typo in the question

username1738583

OP

Original post by Zacken

Probably just the answers being wrong as usual, anyway. When x=3, dy/dx is 6. So that's part (a).

wait so that's it? That's the answer?

Original post by homeland.lsw

wait so that's it? That's the answer?

Yeah.

:tongue:

username1738583

OP

K well this is another one...yuck

The sketch shows the curve

y = x^2 - 5

and the normal at

(2 , -6)

a) Find the equation of the normal at the point

(2 , -6)

So I did
y=x^2 -5
dy/dx = 2x

2x=0
2(2)= 4 (The two coming from the given point)

so 4 is the gradient of the tangent (?)
Then -1/4 is the gradient of the normal (negative reciprocal)

So y= -1/4x + c
-6= -1/4 (2) + c
-11/2= c

Therefore
y= -1/2x + 11/2 for part a?

@Zacken

Original post by homeland.lsw

y= -1/2x + 11/2 for part a?

@Zacken

Is the bolded bit a typo? If so, looks good other than that! Well done. :smile:

:smile:

username1738583

OP

Original post by Zacken

Is the bolded bit a typo? If so, looks good other than that! Well done. :smile:

:smile:

Ah yes it should be -1/4 x for the gradient... :colondollar:

:colondollar:

But then why does the answer say
y=x-8...that's really confusing me :angry:

:angry:

Original post by homeland.lsw

Ah yes it should be -1/4 x for the gradient... :colondollar:

:colondollar:

But then why does the answer say
y=x-8...that's really confusing me :angry:

:angry:

Sounds like whatever answers you're looking at are ****ed to hell. :lol:

:lol:

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