C4 doubts

Did u always want to be a Maths teacher or did u decide it later on in life?

Well it hasn't made the slightest difference to my life at all. I did put Cambridge as my second choice for postgraduate teacher training but I was offered my first choice (the local provider) and this made sense as my children were settled in school.

I was an Independent Financial Adviser for 13 years.

Well, what's the volume after 5 seconds?

Once you've found that, you need to find h by rearranging and substitute to the V formula. Now, find

$\dfrac{dV}{dh}$

Hint:

$\dfrac{dV}{dt} = \dfrac{dV}{dh} \times \dfrac{dh}{dt}$

You're looking for dh/dt, which is the rate of change of the depth.

Hello , pls can u help me with this question (question no 1) on this solomon worksheet
http://www.london-oratory.org/maths/Solomon%20C4%20Worksheets/C4Edex/pdf_files/E4Qint_P.pdf

I don't know what the numbers on the top and bottom of the integral sign would be for this question? so any advice would be greatly appreciated! thanks!

What's the value of x when y=3?

So what are your limits?

But now you need the area of the remainder of the shaded bit.

Hint: it's a rectangle!

Can you do it now?

What's the value of x when y=3?

So what are your limits?

But now you need the area of the remainder of the shaded bit.

Hint: it's a rectangle!

Can you do it now?

So are the numbers which go on top and bottom of the integral sign meant to be 3 and 0?

Noooooo!!

So you know the value of x when y=3? Drop a line from that point on the curve to the x axis, and label the point.

So those are your limits.

But that's not it! You need to add on the area of the remainder of the shaded bit. This remainder is a rectangle!

Hello thanks yes , i can do it now!
Just one more question : is there any easy method or tip i can use to determine which number goes on the top and bottom for questions? what do I need to look out for ? are we looking for the x number to the left and to the right of the enclosed area? and also I was always wondering : doesn't it actually give the area of the whole of under the curve for that portion (i mean right down) why does it stop at the coordinate axes for x?

If you're given x and y in parametric form, i.e

$x=f(t), y=f(t)$

then the value of t that corresponds to the larger value of x will be at the top.

There shouldn't be any confusion otherwise, but be careful with examples like the one you just did. Be sure to annotate the diagram!

To answer your second Q (which I've not seen/heard anyone asking so far), the area does not go right down. Integration just adds up several tiny strips. The width of each tiny strip is dx, a small value in the x direction, and the height is y (the y coordinate).

Thus

$\displaystyle \int_a^b y \ dx$

gives the area of the shape formed by the x axis and the curve y=f(x).

The area all the way down is infinite. Similarly,

$\displaystyle \int_0^b \dfrac{1}{x} \ dx$

where b > 0, is infinite. This is because the y axis is an asymptote.

In some cases, we can integrate over an asymptote, but this is degree level stuff.

If you're given x and y in parametric form, i.e

$x=f(t), y=f(t)$

then the value of t that corresponds to the larger value of x will be at the top.

There shouldn't be any confusion otherwise, but be careful with examples like the one you just did. Be sure to annotate the diagram!

To answer your second Q (which I've not seen/heard anyone asking so far), the area does not go right down. Integration just adds up several tiny strips. The width of each tiny strip is dx, a small value in the x direction, and the height is y (the y coordinate).

Thus

$\displaystyle \int_a^b y \ dx$

gives the area of the shape formed by the x axis and the curve y=f(x).

The area all the way down is infinite. Similarly,

$\displaystyle \int_0^b \dfrac{1}{x} \ dx$

where b > 0, is infinite. This is because the y axis is an asymptote.

In some cases, we can integrate over an asymptote, but this is degree level stuff.

Hello there,
How are you ?
Thanks for taking the time and effort to answer my question in so much detail ! I understand it now ! I was just wondering about these sorts of loopy graphs (like an infinity shape). Are they different because they have got two two chunks (for example: if we are asked to find out the area of the right hand side 'blob' would we have to multiply our answer by 2 to get the final answer?) Thanks and sorry if my questions are sounding too stupid!

http://www.london-oratory.org/maths/Solomon%20C4%20Worksheets/C4Edex/pdf_files/E4Qdif_E.pdf

Hello would we lose the mark if we didn't write down (at the end of the integral thing), with respect to x (like 'dt' ) ? thanks for all ur help so far! did u have a good easter?

I was an Independent Financial Adviser for 13 years.

Yes, because there's a line of symmetry along the x axis (and the y axis too)



You would, because it doesn't make sense unless you say what you're integrating with respect to.

So the examiner would see

$\displaystyle \int \dfrac{1}{x}$

as something that has more than one meaning.

For example

$\displaystyle \int \dfrac{1}{x} \ dt = \dfrac{t}{x} + C$

Yes, I did. Thanks for asking You?

hello , thanks for ur answer!
Pls can you help me to understand this question (question 9e on here) :

http://www.london-oratory.org/maths/Solomon%20C4%20Worksheets/C4Edex/pdf_files/E4Qint_L.pdf


and here is the mark scheme answer :
http://www.london-oratory.org/maths/Solomon%20C4%20Worksheets/C4Edex/pdf_files/E4Aint_L.pdf

I don't understand
what they are doing in that answer! pls help ???

Where do you get stuck?

the earlier part of the question showed you that $ln(\frac{1+m}{1-m}) = 1-e^{-t}$

The limiting value of m is what happens when t gets very big - in this case the exponential gets very small so the RHS gets closer and close to 1.

Next they raise e to the power of both sides to get rid of the logarithm (if ln x = y then x=e^y), and then they rearrange to get m in terms of e, the base of natural logarithms. Finally they convert from kg to g.

Is that OK?

hello , this is the bit that I dont understand . thanks for the help!

$e^{-t} = \dfrac{1}{e^t}$

What happens as t gets larger?

hmmm. that is what i am not understanding !