A Level Maths Exam Question

Original post by pigeonwarrior

https://www.quora.com/profile/Bravewarrior/p-165324030
Here is the question with its solution. I am stuck on part c. To be honest, I'm quite confused and am not even sure on how to start. Some help would be greatly appreciated!

The derivative (rate of growth) is a quadratic, so what value (N) corresponds to the maximum? Then sub into the basic equation to get the corresponding time.

(edited 10 months ago)

Reply 2

OP

Original post by mqb2766

The derivative (rate of growth) is a quadratic, so what value (N) corresponds to the maximum? Then sub into the basic equation to get the corresponding time.

I'm still a bit confused 😬

Reply 3

Original post by pigeonwarrior

I'm still a bit confused 😬

About what? Can you find the value of N for which the gradient is largest? Its a simple quadratic.

Reply 4

OP

Original post by mqb2766

About what? Can you find the value of N for which the gradient is largest? Its a simple quadratic.

I've got it now, thank you! 🙂

Reply 5

math-path

5

Original post by pigeonwarrior

I'm still a bit confused 😬

On part (b) you found dN/dt and it is a quadratic equation in terms of N. The quadratic has negate N^2 so the curve opens down and it has a maximum point at the vertex. To find the vertex coordinates for N you need to complete the squares and you will find N=150.

Screenshot 2024-05-23 114549.png

(edited 10 months ago)

Reply 6

Original post by math-path

On part (b) you found dN/dt and it is a quadratic equation in terms of N. The quadratic has negate N^2 so the curve opens down and it has a maximum point at the vertex. To find the vertex coordinates for N you need to complete the squares and you will find N=150.
Screenshot 2024-05-23 114549.png

Tbh, I suspect they simply want you to note its 300/2, by symmetry, as the roots are 0 and 300. Not that different from completing the square but completing the square is a bit overkill to do it as you just need N, not the value of the maximum.

Reply 7

OP

Original post by math-path

On part (b) you found dN/dt and it is a quadratic equation in terms of N. The quadratic has negate N^2 so the curve opens down and it has a maximum point at the vertex. To find the vertex coordinates for N you need to complete the squares and you will find N=150.
Screenshot 2024-05-23 114549.png

Thank you so much! 🙂

Reply 8

OP

Original post by mqb2766

Tbh, I suspect they simply want you to note its 300/2, by symmetry, as the roots are 0 and 300. Not that different from completing the square but completing the square is a bit overkill to do it as you just need N, not the value of the maximum.

Wait how did you know the roots are 0 and 300?

Reply 9

OP

Original post by pigeonwarrior

Wait how did you know the roots are 0 and 300?

Nevermind I guess you would make the derivative = 0 🫢
My bad

Reply 10

Original post by pigeonwarrior

Nevermind I guess you would make the derivative = 0 🫢
My bad

No. Its given to you in factorised from so
N(300-N)
So roots of 0 and 300 and the max at the midpoint. Hence the "deduce" in the mark scheme.

Reply 11

OP

Original post by mqb2766

No. Its given to you in factorised from so
N(300-N)
So roots of 0 and 300 and the max at the midpoint. Hence the "deduce" in the mark scheme.

Yup, that's what I meant, I just badly worded it 😬 I meant that by making dN/dt=0 means the numerator is 0. So N=0 and 300. And dN/dt has to be zero where the roots are i think which is why you would equate it to 0 in the first place

Reply 12

ClineRob