Sum or difference of 7/8 + 1/3
2 answers:
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Let 
be an arbitrary point on the surface. The distance between 
and the given point 
is given by the function
Note that
and 
attain their extrema, if they have any, at the same values of 
. This allows us to consider the modified distance function,
So now you're minimizing
subject to the constraint 
. This is a perfect candidate for applying the method of Lagrange multipliers.
The Lagrangian in this case would be
which has partial derivatives
Setting all four equation equal to 0, you find from the third equation that either
or 
. In the first case, you arrive at a possible critical point of 
. In the second, plugging into the first two equations gives
and plugging these into the last equation gives
So you have three potential points to check:, 
, and 
. Evaluating either distance function (I use 
), you find that
So the two points on the surface closest to the point are 
.
Note that
So now you're minimizing
The Lagrangian in this case would be
which has partial derivatives
Setting all four equation equal to 0, you find from the third equation that either
and plugging these into the last equation gives
So you have three potential points to check:
So the two points on the surface
Answer: t-half = ln(2) / λ ≈ 0.693 / λ
Explanation:
The question is incomplete, so I did some research and found the complete question in internet.
The complete question is:
Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:
1) Equation given:
← I used α instead of λ just for editing facility..
Where No is the initial number of nuclei.
2) Half of the initial number of nuclei: N (t-half) = No / 2
So, replace in the given equation:
3) Solving for α (remember α is λ)
αt ≈ 0.693
⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ
Explanation:
The question is incomplete, so I did some research and found the complete question in internet.
The complete question is:
Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:
N(t)=No e−λt,
where λ is known as the decay constant. Note that at t=0, N(t)=No, the original number of unstable nuclei. N(t) decreases exponentially with time, and as t approaches infinity, the number of unstable nuclei that remain approaches zero.
Part (A) Since at t=0, N(t)=No, and at t=∞, N(t)=0, there must be some time between zero and infinity at which exactly half of the original number of nuclei remain. Find an expression for this time, t half.
Express your answer in terms of N0 and/or λ.
Answer:1) Equation given:
Where No is the initial number of nuclei.
2) Half of the initial number of nuclei: N (t-half) = No / 2
So, replace in the given equation:
3) Solving for α (remember α is λ)
αt ≈ 0.693
⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ