Mathematics Question Thread

[anotherworld2b]

thank you for your help
I was wondering for the base point what would you do if it changes from 25 to 26? Would the base point be -1? (25-26?)

Since you've started the 1st one, I'll show you the full working of the 2nd and hopefully it will help you finish the first!

Remember, the formula we use is:



This comes from rearranging the visually simpler approximation, \(\frac{\delta C}{\delta n}\approx\frac{dC}{dn}\) - Remember that the small change (on the left) is roughly equal to the derivative (on the right), when that derivative is evaluated at the correct point. Geometrically, this is the same as using a tangent to approximate a curve (as shadow said):



We also know that \(\delta n=1\), so:



So the additional cost will be $200, roughly. If you compare that to the actual cost found by substitution, it is $216, so we're reasonably close

So the steps are to find the derivative, and evaluate it at our base point (what we're increasing from), then use the formula

[anotherworld2b]

I still dont really understand the steps to solve it 

[Shadowxo]

If it changes from 25 to 26:
The derivative at a certain point= dy/dx = gradient at that point= change in y divided by change in x (approximation, if that gradient stayed the same) = change in y per x
So you'd find f'(x), aka the gradient, sub in x=25
f'(25) = change in y divided by change in x = ∆y/∆x
x increases by 1 (x2-x1=1)
So f'(25)=∆y/1
∆y=f'(25)
This change in y is the approximation if the gradient were to stay the same from that point - like drawing a tangent at a point and using that to approximate the increase in y

[smile123]

PLEASE HELP

[Shadowxo]

I still dont really understand the steps to solve it 

Remember, pi is a constant, not the variable r
So A'(r) = 2πr - only one step needed (also you should put it in terms of r : A'(r) instead of A'(x))
So where A=120, r=6.180
dA/dr = 2*π*6.180
Change in A = 1
Change in r : unknown
dA/dr = 1/∆r = 2*π*6.180
∆r = 1/(2π*6.180)

Hope this helps

[Shadowxo]

PLEASE HELP

|cos(2x)|=1
cos(2x) =1 or -1
2x = 0,π,2π,3π,4π,5π... (as cos of those values equals 1 or -1)
x = 0, π/2, π, 3π/2, 2π...
5 solutions within the domain
Let me know if you want me to explain anything further

[anotherworld2b]

I am bit confused what to do for percentages

[Shadowxo]

I am bit confused what to do for percentages

This is a bit of a tough question, and I'm not sure whether that would usually be tested. Also, remember you're differentiating A with respect to w (dA/dw) and k is a constant so you don't need to use the product rule to differentiate - treat is as though it is a number, like 4 or 7.
Anyway, here's the solution:

[Fahim486]

Hi everyone,
Can someone pls help me with question 9 b) and c)

Thanks!

[Fahim486]

Sorry I also need help with this question too

Thanks again!!

[Shadowxo]

9. So we have V =4x³-220x²+2400x
V is a max when dV/dx = 0
dV/dx = 12x²-440x+2400
12x²-440x+2400=0
Solve for x
for c) just substitute in the x value you found into V

5. We know that the rest of the perimeter, (not including the walls) equals 8
L+w=8 -> w=8-L
A = L*w = L*(8-L) = 8L-L²
Max area when dA/dL=0
dA/dL = 8-2L
8-2L=0
L = 4
w=8-L = 4
Max area when length =4, width =4

[laurenf58]

Can someone please remind me how to do this? Thanks!

[RuiAce]

Can someone please remind me how to do this? Thanks!

[smile123]

PLEASE HELP

[RuiAce]

PLEASE HELP