2u Mathematics Marathon v1.0 (1 Viewer)

[Jono_2007]

A radioactive substance is decaying at a rate proportional to its mass. Its mass initially is measured at 37.4 grams. Twelve hours later the mass is measured at 31.7 grams. Find:

(i) the decay constant
(ii) the half-life of the substance

Spoiler

In(N/N₀₎=k.t

i)In (31.7/37.4)=12k
-0.165354023=12k
:.k=-0.01377950 hrs^-1

ii)N=N₀.e^-k. t
(1/2)N=N₀.e^-0.01377950. t
(1/2)=e^-0.01377950. t
In (1/2)=In. (e^-0.01377950)
In (1/2)=-0.01377950. t
:. t=In (1/2)/(-0.01377950)

therefore the halflife is 50.302 hours

 

[SoulSearcher]

Congratulations! You must know your basics well then! Go for 4 unit!

thank you

not sure about doing 4 unit as yet, just concentrating on extension 1 for now

 

[SoulSearcher]

Spoiler

A=a₀e^kt

ii) 31.7=e^12k

In 31.7=In e^12k

In31.7=12k

:.k=(In 31.7)/12

decimal apoximation is k=0.288;

1/2A₀=A₀e^0.288 t

1/2=e^0.288 t

In (1/2)=In [e^0.288 t]

In(1/2)=0.288 t

:.t=In (1/2)/(0.288)

:. the half life is 2.407 days

Spoiler

in the line in bold you forgot 2 things:

1. where is the original amount i.e. 37.4
2. as this is exponential decay, the value of k has to be negative, therefore -12k

 

[SoulSearcher]

ill give the answer tomorrow if no one else has figured it out by then

 

[SoulSearcher]

Spoiler

In(N/N₀₎=k.t

i)In (31.7/37.4)=12k
-0.165354023=12k
:.k=-0.01377950 hrs^-1

ii)N=N₀.e^-k. t
(1/2)N=N₀.e^-0.01377950. t
(1/2)=e^-0.01377950. t
In (1/2)=In. (e^-0.01377950)
In (1/2)=-0.01377950. t
:. t=In (1/2)/(-0.01377950)

therefore the halflife is 50.302 hours

that is correct

you can post the next question

 

[Jono_2007]

Next question:
Two straight roads Ab and AC, are inclined to each other at 54 degrees. Two bike riders begin simultaniosly from A and travel along the roads at 16 and 24 km.h^-1 respectively. How long is it before they are 60 km apart in a direct line? Answer to the nearest minute.

Enjoy!

 

[icycloud]

Next question:
Two straight roads Ab and AC, are inclined to each other at 54 degrees. Two bike riders begin simultaniosly from A and travel along the roads at 16 and 24 km.h^-1 respectively. How long is it before they are 60 km apart in a direct line? Answer to the nearest minute.

Spoiler

Observe that distance = speed x time.

Now, let
d(AB) = s(AB) * t(AB) = 16 * t(AB)
d(AC) = s(AC) * t(AC) = 24 * t(AC)

But, t(AB) = t(AC) [they travel for the same period of time], so let T = t(AB) = t(AC)

Thus, we have d(AB) = 16t, d(AC) = 24t
Now, using the cosine rule, we have:

Cos(54°) = [(16t)^2 + (24t)^2 - 60^2] / [2 . 16t . 24t]
768Cos(54°) t^2 = 832t^2 - 3600
t = +/- Sqrt( -3600 / (768Cos(54°) - 832))
= +/- 3.07558...

Time is always positive, thus we take the positive answer.
t = 3.07558...
= 3 hours 4 minutes 32.11 seconds
= 3 hours 5 minutes (nearest minute)
#

Next question:
Find ∫cos(x)ln[sin(x)^e] dx {Using the substitution u = sin(x)}
First by differentiating xln(x)

 

[Jono_2007]

I've tried teaching myself integration but i dont know where to start, like what was the first part of integration that u guys learnt. (antideriviative/indefinte integral) or what!

 

[Riviet]

Riv's integration tutorial for beginners:

"Find the primitive", "Find the anti-derivative", and "Integrate" mean the same thing.

To integrate y=axⁿ, add 1 to the index and divide by this new index.

So the primitive of xⁿ = axⁿ⁺¹ / n+1

Try these out.

Find the primitive of:

a) x²

b) 2x

c) x³ / 3

d) 1 [hint: 1= 1.x⁰]

e) x^1/2

How's that?

 

[Jono_2007]

Cheers Riviet!

 

[Jono_2007]

So if the primitive of X^n is x^n+1/n+1; then the dirivative of x^n is dy/dx=n.x^n-1 is that right. But what does with respect to x mean, i've seen it alot.

 

[Riviet]

So if the primitive of X^n is x^n+1/n+1; then the dirivative of x^n is dy/dx=n.x^n-1 is that right. But what does with respect to x mean, i've seen it alot.

Yep you got it.

"With respect to x" just means to integrate/differentiate something as a function of x, so integrating y=f(x) is integrating y with respect to x.

 

[insert-username]

Next question:
Find ∫cos(x)ln[sin(x)^e] dx {Using the substitution u = sin(x)}
First by differentiating xln(x)

Integration through substitution is 3-unit, isn't it?


I_F

 

[Riviet]

Hmm... in the hsc 2 unit, I think substitution is in it.

 

[SoulSearcher]

im going to try this but i have no idea if this is right

Next question:
Find ∫cos(x)ln[sin(x)^e] dx {Using the substitution u = sin(x)}
First by differentiating xln(x)

Spoiler

d/dx {x ln (x)} = ln(x) + 1

therefore

x ln (x) + c = ∫ {ln (x) + 1} dx
x ln (x) + c = ∫ ln (x) dx + ∫ 1 dx
therefore
x ln (x) + c = x + ∫ ln (x) dx
and
x ln (x) - x + c = ∫ ln (x) dx
therefore we can substitute u for x and thus
u ln (u) - u + c = ∫ ln (u) du

now
∫ cos (x) ln[sin(x)^e] dx
= ∫ cos (x) dx * ∫ ln[sin(x)^e] dx
= ∫ cos (x) dx * e ∫ ln (sin x) dx
= ∫ cos (x) dx * e ∫ ln (u) du , where u = sin (x)

now substituting u ln (u) - u for ∫ ln (u) du
∫ cos (x) dx * e ∫ ln (u) du
= ∫ cos (x) dx * e[u ln (u) - u], disregarding the constant
= {sin (x) * e[u ln (u) - u] } + C
= { u * e[u ln (u) - u] } + C, again using the substitution u = sin (x)
= { e[u² ln (u) - u²] } + C
= { e [sin² (x) ln (sin x) - sin² (x)] } + C

i am not sure if this is right, indeed i have no idea what i just did, so if you can check this is right it would be good

 

[Riviet]

Next question:
Find ∫cos(x)ln[sin(x)^e] dx {Using the substitution u = sin(x)}
First by differentiating xln(x)

I think this is right :

Spoiler

d/du(ulnu)=lnu+1

Integrating LHS and RHS,

ulnu=(∫lnu du) + u

∫lnu du=ulnu-u (1)

Now let u=sinx

du/dx=cosx

du=cosx dx

.: ∫cos(x)ln[(sinx)^e] dx =∫lnu^e du

=e∫lnu du

=e(ulnu-u)+C, from (1)

=esinx(ln[sinx]-1) + C

Next Question:

This is a fun one that everyone can have a go at

If x²+y²=14xy, express in terms of log x and log y:

a) log (x+y)

b) log (x-y)

c) log (x²-y²)

 

[KeypadSDM]

If x²+y²=14xy, express in terms of log x and log y:

a) log (x+y)

b) log (x-y)

c) log (x²-y²)

Spoiler

x²+y²=14xy
a) Add 2xy to both sides
(x + y)² = 16xy
2Log[x + y] = Log[2⁴] + Log[x] + Log[y]
Log[x + y] = 1/2(Log[x] + Log[y]) + 2Log[2]

b) Subtract 2xy from both sides:
(x - y)² = 12xy
2Log[x - y] = 2Log[2] + Log[3] + Log[x] + Log[y]
Log[x - y] = Log[2] + 1/2(Log[x] + Log[y] + Log[3])

c) Log[x²-y²]
= Log[x - y] + Log[x + y]
= Log[2] + 1/2(Log[x] + Log[y] + Log[3]) + 1/2(Log[x] + Log[y]) + 2Log[2]
= Log[x] + Log[y] + 3Log[2] + (1/2)Log[3]

EDIT: I did a stupid error, god I'm dumb.

I have to admit, that was quite "fun". For a maths question.

 

[Riviet]

Indeed, that's a fun one.

Keypad, would you like to post the next question?

 

[KeypadSDM]

A simple one, just because I'm nice:

∫tanⁿ[x]dn

 

[icycloud]

A simple one, just because I'm nice:

∫tanⁿ[x]dn

Spoiler

I=∫tanⁿ[x]dn

Let c = tan[x] (constant)

Now I = ∫cⁿ dn
= ∫e^(nln(c)) dn
= 1/ln(c) ∫ln(c) e^(nln(c)) dn
= cⁿ/ln(c) + C
= tanⁿ[x] / ln(tan[x]) + C
#

Next question:
A and B are two points 200 metres apart. For what values of L is it possible to find a circular arc AB of length L metres? Justify your answer.