A block of radioactive isotope is stored in a laboratory, given the following statistics:

1.The isotope have a half life of exactly 3 years. (A year is equal to 31,500,000 seconds in this question.)

2.The radiation power is \(100W/m^2\) 1 meter away from the block when it is placed in the lab.

3.With every half life (3 years) passed, the radiation would have decreased by a factor of two (hence the name "half")

4.Radiation power is proportional to the reverse square of distance to the block. The radiation power of the block can be given by the following function, with \(x\) being the distance away (In meters) and \(y\) being the power (In Watts):

\(y=\frac{100}{x^2},\space x>0\)

Given the stats above, calculate:

The total radiation power in Joules \((J)\) received by a steel plate with area \(9m^2\) and \(5\) meters away from the beginning of third year to the middle of the eighth year from the block's placement, give your answer in scientific notations.)

(This question requires moderate knowledge in integral calculus to be solved.)

(Bet a hundred dollar this question will still be unanswered in the morning.)

(Well I didn't expect for someone to actually answer this XD)

Jeffes02
Aug 18, 2017

#1**+3 **

Try not to confuse power (Watts), power density (Watts per square meter) and Energy (Joules)!

Here's my interpretation:

.

Alan
Aug 18, 2017

#3**+2 **

Corrections: (i) I should have set t_{1} to 2 years not 2/3 years! The result is then E = 2.224*10^{9} J

(ii) The expression for E (after integration) should have t_{1} and t_{2} interchanged (or the result will be a negative energy!).

.

Alan
Aug 18, 2017

#4**-2 **

I can't quite understand your process, not being offensive but it looks a bit messy. (I prefer LaTeX more XD)

Checking your answer:\(\int _2^{7.5}3^{-x}\cdot\frac{100}{5^2}\cdot9\cdot31,500,000\space dx\)

Simplify & Pull the constant out:

\(=1,134,000,000\int _2^{7.5}3^{-x}dx\)

Now solving: \(\int3^{-x}dx\)

Substitute \(u=-x,\space dx=-du\)

\(=-\int3^udu\)

Apply exponential rule:

\(=-\frac{3^u}{\ln \left(3\right)}\)

Unsubstitude \(u\):

\(-\frac{3^{-x}}{\ln \left(3\right)}\)

So: \(\int3^{-x}dx=-\frac{3^{-x}}{\ln \left(3\right)}\)

Therefore: \(\int _2^{7.5}3^{-x}dx=\left(-\frac{3^{-7.5}}{\ln \left(3\right)}\right)-\left(-\frac{3^{-2}}{\ln \left(3\right)}\right)\)

\(\approx0.1009\)

\(Ans=0.1009\cdot1,134,000,000\)

\(\approx114420000 (Joules)\)

(Wait a minute, why is my answer different from yours @_@)

Jeffes02
Aug 19, 2017

#5**+1 **

**Not to be offensive or anything Jeffes but I think your response to Alan's answer is extremely offensive!**

Whatever happened to "Thanks you very much, I really appreciate all you time and expertise"!!

Thanks Alan, that is a personal thank you from me :D

Melody
Aug 19, 2017

#6**0 **

Jeff: Please do not be arrogant in trying to be a "show-off". As Melody says, you should be polite and thankful towards your elders.

Guest Aug 19, 2017

#2**+1 **

Jeff, be sure to send Alan his $100.LOOOOL (Bitcoin should should work just fine).

GingerAle
Aug 18, 2017

#8**+1 **

Alan is right !!!.

Compute the definite integral:

1134000000 integral_2^7.5 2^(-x/3) dx

For the integrand 2^(-x/3), substitute u = -x/3 and du = -1/3 dx.

This gives a new lower bound u = -2/3 = -2/3 and upper bound u = -7.5/3 = -2.5:

= -3402000000 integral_(-2/3)^(-2.5) 2^u du

Switch the order of the integration bounds of 2^u so that the upper bound is larger. Multiply the integrand by -1:

= 3402000000 integral_(-2.5)^(-2/3) 2^u du

Apply the fundamental theorem of calculus.

The antiderivative of 2^u is 2^u/(log(2)):

= (26578125 2^(u + 7))/(log(2)) right bracketing bar _(-2.5)^(-2/3)

Evaluate the antiderivative at the limits and subtract.

(26578125 2^(u + 7))/(log(2)) right bracketing bar _(-2.5)^(-2/3) = (26578125 2^(7 - 2/3))/(log(2)) - (26578125 2^(7 - 2.5))/(log(2)) = 2.22425×10^9:

**Answer: | = 2.22425×10^9 Joules.**

Guest Aug 19, 2017

#10**0 **

Jeff, **this is an example of messy: ** Complex mathematics presented in ASCII text.

Mr. BB presentations as the *Answer Man *are usually just **a** **plop of slop** –a vast sea of ASCII soup coupled with descriptions of mathematical symbols pasted and plastered from Wolfram Alpha’s automated solutions. I usually need a double dose of Dramamine to prevent seasickness when I try to decipher them. This particular presentation is a little better than his usual slop. This still requires wading, but you just need wading boot instead of the usual hip boots. I didn’t need the Dramamine this time.

Actually, I’m impressed that Mr. BB can query Wolfram proficiently enough to obtain this answer. Aside from Wolfram’s claim for natural language query, higher-level math usually requires precise wording or symbolic code to prevent absurd interpretations.

I was going to teach my cat how to do this, but my dog volunteered to teach him for me. (Good dog, Mr. Peabody!)

GingerAle
Aug 19, 2017

#9**+1 **

\(\text {Sir Alan's presentation in LaTeX}\\ \tiny \text {Reproduced verbatim (as much as practicable) with corrections from listed errata. }\\ \text { }\\ \small \text {definition: } \hspace{25 mm} years : = 315000000 {{\mkern 1mu\cdot\mkern 1mu}} s\\ \small \text {half-life: } \hspace{28 mm} \tau: =3{{\mkern 1mu\cdot\mkern 1mu}} years\\ \small \text {initial isotope power: } \hspace{6 mm} Q_0=100{{\mkern 1mu\cdot\mkern 1mu}}W\\ \text{ }\\ \small \text {isotope power as } \\ \small \text {a function of time: } \hspace{10 mm} Q(t): = Q_0{{\mkern 1mu\cdot\mkern 1mu}}e^{-ln(2)\cdot(\frac{t}{\tau})}\\ \small \text {plate area: } \hspace{22 mm} A: = 9{{\mkern 1mu\cdot\mkern 1mu}} mm^2\\ \small \text {plate distance: }\hspace{16 mm} x: = 5 {{\mkern 1mu\cdot\mkern 1mu}} mm\\ \)

\(\small \text {power received at plate } \\ \small \text {as a function of time: } \hspace{9 mm} P(t): = \frac{A \cdot Q(t)}{x^2}\\ \small \text {beginning of third year: } \hspace{9 mm} t_1: = 3 {{\mkern 1mu\cdot\mkern 1mu}} years\\ \small \text {middle of eighth year: } \hspace{12 mm} t_2: = 7.5\; years\\ \small \text { }\\ \small \text {total ENERGY received: } E:= \int_{t_1}^{t_2} P(t) \; dt\\ \text{ }\\ \hspace{55 mm} E=\frac{A}{x^2}\cdot Q_0 \cdot \frac{\tau}{ln(2)} \cdot (e^{\frac{ln(2)\cdot t_2}{\tau}} -e^{\frac{ln(2)\cdot t_2}{\tau}}) \text{ }\\ \hspace{55 mm} E = 2.224 \cdot 10^9 J\\\)

-----------

By reproducing this, it seems I’ve demonstrated two proofs. Q.E.D.

1) If you prefer the format in LaTeX, you can do it yourself. It has the same *messiness *but with a different font. For an example of true messiness, see my troll post for the “Answer Man,” aka Mr. BB.

2) **You are an idiot on multiple levels!** You’ve risk alienating the only person (in regular attendance) who can answer this type of question. You are well too far on the wrong side of the education gulf to be this arrogant. If you continue, **you will only ever earn a dimwit degree in Stupid. **

Mr. BB has a dimwit degree in Stupid. He is about as educated as you are but he’s more arrogant. He uses his degree as a license to torture forum members with presentations so incompetent that it amuses my cat. My cat likes him for some reason. I really don’t know why. Maybe because he makes him laugh or maybe because my cat is a (Monopoly) banker and embezzler, or maybe it’s because he likes rodents. It’s probably all three.

GingerAle
Aug 19, 2017