Random math questions

[lzxnl]

How do you evaluate ? Tried L'hopital's rule but ended with something like and differentiating top and bottom just gives pretty much the same thing

If you take out a factor of 3^x from the bracket, you get (1+(2/3)^x)^(1/x)×3. The bracket approaches 1 and the power approaches 0 so your final limit is 3.
Alternatively, your function is bounded between (3^x+0)^(1/x) and (3^x+3^x)^(1/x). Either of those limits should be easier to do.

L hopital's rule doesn't solve all limits unfortunately.

[#1procrastinator]

Thanks, brah!

[idontknow2298]

If the edges of a rectangular box were increased by 1cm, 2cm and 3cm respectively, the box would become a cube and its capacity would be increased by 101 cubic cm. Find the dimensions of the rectangular box.
How would I go about doing this question?

[RKTR]

If the edges of a rectangular box were increased by 1cm, 2cm and 3cm respectively, the box would become a cube and its capacity would be increased by 101 cubic cm. Find the dimensions of the rectangular box.
How would I go about doing this question?

let sides of cube=x ,volume of cube = x^3

dimensions of rectangular box = x-1,x-2,x-3  , volume of rectangular box = (x-1)(x-2)(x-3) = x^3-6x^2+11x-6

V of cube - V of rectangular box =101
6x^2-11x+6=101
6x^2-11x-95=0
(6x+19)(x-5)=0
x=-19/6 or x=5
x>0 therefore x=5
dimensions of rectangular box =4 cm , 3 cm , 2 cm

[idontknow2298]

Thanks!

[idontknow2298]

How do you do this question?
The volume of a solid sphere varies directly as the cube of its radius. The building material of three spheres with radii 3 cm, 4 cm and 5 cm is melted and a new sphere is then made. Find the radius of this new sphere.

[idontknow2298]

How about this one:
Give that y varies directly with (ax-2) find the value of a if x=3 and y=2; and if x=4 and y=4

[#1procrastinator]

^ just plug in the given values of x and y and solve for a


how do you get from antidifferentiating   i'm getting from substitution + partial fractions but the plot doesn't fully agree with the solution wolframalpha gave

[#1procrastinator]

Is the intersection of finitely many closed sets closed? I'm aware it is for a 'collections of sets' but I'm not sure if this means that it has to be an infinite collection or it can be finite.
In general, if something is stated for an indexed collection of sets, does it mean it applies to both finitely many and infinitely many sets?

[lzxnl]

^ just plug in the given values of x and y and solve for a


how do you get from antidifferentiating   i'm getting from substitution + partial fractions but the plot doesn't fully agree with the solution wolframalpha gave

Erm...I don't think I quite understand your answer. I don't think you're meant to get a spare (1+cos x) term anywhere. Is that under another log or something?

dx/sin x => sin x dx/(1-cos^2 x)
Let u = cos x, du = -sin x dx
Integral becomes du/(u^2-1) => du/2 (1/(u-1) - 1/(u+1)) => 1/2 ln(1-cos x) - 1/2 ln(1+cos x) after integrating and back subbing
That's what you meant, right?

OR you can do this
1/(sin x)
Sub t = tan x/2 (dirtiest substitution trick in the book)
You can show quite easily that tan x = 2t/(1-t^2), sin x = 2t/(1+t^2) and cos x = (1-t^2)/(1+t^2) , dx = 2 dt/(1+t^2)
so our integral becomes (1+t^2)/2t * 2dt/(1+t^2)
The 1+t^2 terms drop off
so our integral is just dt/t
We're left with just ln (tan x/2) which really should appear as ln(sin (x/2)) - ln(cos(x/2))

[#1procrastinator]

Erm...I don't think I quite understand your answer. I don't think you're meant to get a spare (1+cos x) term anywhere. Is that under another log or something?

dx/sin x => sin x dx/(1-cos^2 x)
Let u = cos x, du = -sin x dx
Integral becomes du/(u^2-1) => du/2 (1/(u-1) - 1/(u+1)) => 1/2 ln(1-cos x) - 1/2 ln(1+cos x) after integrating and back subbing
That's what you meant, right?

Yeah, typo - was in a hurry when I was writing that


OR you can do this
1/(sin x)
Sub t = tan x/2 (dirtiest substitution trick in the book)
You can show quite easily that tan x = 2t/(1-t^2), sin x = 2t/(1+t^2) and cos x = (1-t^2)/(1+t^2) , dx = 2 dt/(1+t^2)
so our integral becomes (1+t^2)/2t * 2dt/(1+t^2)
The 1+t^2 terms drop off
so our integral is just dt/t
We're left with just ln (tan x/2) which really should appear as ln(sin (x/2)) - ln(cos(x/2))

'World's sneakiest substitution' haha. How come the two aren't 'exactly' the same though?
The bold plot is of the function obtained from the dirty sub.

[lzxnl]

Try putting the absolute value signs now in the logs. I don't think I did that.

[#1procrastinator]

ah yeah  thanks lxznl
---

Is the intersection of finitely many closed sets closed? I'm aware it is for a 'collections of sets' but I'm not sure if this means that it has to be an infinite collection or it can be finite.
In general, if something is stated for an indexed collection of sets, does it mean it applies to both finitely many and infinitely many sets?

[idontknow2298]

Can someone please help with this question:
If z varies directly as x when y is constant and inversely as y when x is constant and if z=8 when x=4 and y=2, find z when x=3 and y= 1/3

[kinslayer]

Is the intersection of finitely many closed sets closed? I'm aware it is for a 'collections of sets' but I'm not sure if this means that it has to be an infinite collection or it can be finite.
In general, if something is stated for an indexed collection of sets, does it mean it applies to both finitely many and infinitely many sets?

An intersection of arbitrarily many closed sets is closed. A union of arbitrarily many open sets is open.

An intersection of finitely many open sets is open, while a union of finitely many closed sets is closed.

An interesection of infinitely many open sets need not be open:   converges to . Similar counterexamples exist for an infinite union of closed sets.