Edexcel A2 C4 Mathematics June 2016 - Official Thread

Original post by Craig1998

30+50e^0.002t> 84
50e^0.002t>54
e^0.002t>27/25
t/500>ln1.08
t>500ln1.08
t>38.4

Therefore t = 39
Therefore year = 1980+39
= 2019.

Oh god I'm failing c4. :frown:

Yep thats what I got, but the mark scheme says 2018, must be an error on their part.

Btw this is C3 :cute:

, im sure you'll be fine. Thanks!

Reply 3042

Q) Why is the gradient of tangent --> 1/(value of dy/dx)

when https://gyazo.com/1cd6bd9ac94dbec89e08132a94ff86a1
shows the gradient of tangent is the exact value of dy/dx yet this shows the gradient of tangent
is https://gyazo.com/a31c58f4855ac1c0d1d8b6177b8a08a2 .

I understand when you draw a picture you can see gradient of tangent is 1/gradient but say for example
if they say ' find the equation of tangent of the line. ' then the tangent 'm' value in the equation would be
the dy/dx value and if they ask 'find the equation of tangent TO A PARTICULAR POINT' then do you do 1/gradient. <--- is that right? :sigh:

I don't know why this has confused me last minute :lol:

(edited 7 years ago)

Reply 3043

Zacken

22

wot. Gradient of tangent is dy/dx.

Reply 3044

Original post by Zacken

wot. Gradient of tangent is dy/dx.

That's what I thought. I was confused by what that person did in that workings. :confused:

Can you explain this to me please.

Question: https://gyazo.com/dbdd9ce6a296f8c337e633e69cd442b3

Mark scheme: https://gyazo.com/0a971e1e6c6cd570288ad4b35f6aae62

I got that 1st line in the mark scheme where I dy/dx: 3sqroot(2) . cos (theta+ alpha)

and dy/dx = 1/2 so why can't I do 3sqroot(2) . cos(theta+alpha) = 1/2?

How did they get -2 or 2? Is it from the graph or something I missed? :erm:

Reply 3045

Zacken

22

Original post by XxKingSniprxX

...

...you could have given us the question in the first place.

This is of the form x = f(y). So the gradient of the tangent is given by dx/dy.

In most questions, the curve is defined as y = f(x) so the gradient of the tangent is dy/dx.

In the question you've linked, you've found x = f(y), so by differentiating you get dx/dy = f ' (y).

You want dy/dx. So dx/dy = 1/(dy/dx).

Reply 3046

Original post by Zacken

...you could have given us the question in the first place.

This is of the form x = f(y). So the gradient of the tangent is given by dx/dy.

In most questions, the curve is defined as y = f(x) so the gradient of the tangent is dy/dx.

In the question you've linked, you've found x = f(y), so by differentiating you get dx/dy = f ' (y).

You want dy/dx. So dx/dy = 1/(dy/dx).

I've been reading the question as y = f(x) ---> L :frown:

Thanks, I understood it easy once I realised it was x = f(y)

Reply 3047

Zacken

22

Original post by XxKingSniprxX

I've been reading the question as y = f(x) ---> L :frown:

Thanks, I understood it easy once I realised it was x = f(y)

Coolio.

Reply 3048

yesyesyesno

3

Original post by XxKingSniprxX

Q) Why is the gradient of tangent --> 1/(value of dy/dx)

when https://gyazo.com/1cd6bd9ac94dbec89e08132a94ff86a1
shows the gradient of tangent is the exact value of dy/dx yet this shows the gradient of tangent
is https://gyazo.com/a31c58f4855ac1c0d1d8b6177b8a08a2 .

I understand when you draw a picture you can see gradient of tangent is 1/gradient but say for example
if they say ' find the equation of tangent of the line. ' then the tangent 'm' value in the equation would be
the dy/dx value and if they ask 'find the equation of tangent TO A PARTICULAR POINT' then do you do 1/gradient. <--- is that right? :sigh:

I don't know why this has confused me last minute :lol:

I would expect someone getting 95+ to not be thrown off by this of all things. :tongue:

Reply 3049

Original post by yesyesyesno

I would expect someone getting 95+ to not be thrown off by this of all things. :tongue:

Haha.. we all have our :shot:

moments.

Reply 3050

imran_

anyone have the AQA C3 paper? ive heard it was a beast of a paper. I wanna see how hard it really was

Reply 3051

How do I go from my alternative answer to getting the quadratic equation they formed in this question?

https://gyazo.com/861095830fbde070b20076a796aeba94

Reply 3052

ImNervous

Original post by XxKingSniprxX

How do I go from my alternative answer to getting the quadratic equation they formed in this question?

https://gyazo.com/861095830fbde070b20076a796aeba94

That's how you do it. But to be honest I don't know why your way won't work. Always hated that about logs. For some reason I can only get one way to work. Rest give me wrong answers even if they seem right.

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1466382740555.jpg17.2KB

Reply 3053

SaadKaleem

8

Original post by XxKingSniprxX

How do I go from my alternative answer to getting the quadratic equation they formed in this question?

https://gyazo.com/861095830fbde070b20076a796aeba94

You can't antilog like that, you have to use the log rule, from C2: -log(x) is log(1/x)

Once you've both logs on the side, you can anti-log it.

Reply 3054

ImNervous

Original post by SaadKaleem

You can't antilog like that, you have to use the log rule, from C2: -log(x) is log(1/x)

Once you've both logs on the side, you can anti-log it.

Do you mind explaining further please. I don't understand. I get how to do it but I don't get why the way he did it first doesn't work. Please and thanks.

Posted from TSR Mobile

Reply 3055