Question

Help Please!
*METAL
Evaluate the expression:

v ⋅ w

Given the vectors:

r = <5, -5, -2>; v = <2, -8, -8>; w = <-2, 6, -5>

Answer 1

The dot product of two vector u = <a, b, c> and k = <d, e, f> is
\(u \cdot k = ad + be + cf\)

Answer 2

As you can see, the dot product return a scalar. While a cross product will return a vector

Answer 3

oh ok.

Answer 4

2(-2) + (-8)6 + (-8)-5

Answer 5

= -12
Right?

Answer 6

So?

Answer 7

That's correct!

Answer 8

Thank you!

Answer 9

Np ;)

Answer 10

Do you have time for another prob.

Answer 11

sure

Answer 12

A certain radioactive isotope has a half-life of 15 days. If one is to make a table showing the half-life decay of a sample of this isotope from 32 grams to 1 gram; list the time (in days, starting with t = 0) in the first column and the mass remaining (in grams) in the second column, which type of sequence is used in the first column and which type of sequence is used in the second column?

Answer 13

omg, my computer... ugh.

Answer 14

The single element radioactive decay law is:
\(dN/dt = - \lambda N\), where \(N\) is the mass at time t and \(\lambda\) is the decay constant of the element

Answer 15

That's a first order ordinary differential equation of simple solution, just integrate:
\(N(t) = N_0 e^{- \lambda t} \)

Answer 16

Where \(N_0\) is teh initial mass

Answer 17

Well, let's find a ralation of \(\lambda\) with the half-life \(t_{1/2}\)

Answer 18

\(N_0 /2 = N_0 e^{- \lambda t}\), now find t that represents the half-life:
\[t = t_{1/2} = \frac{ln(2)}{\lambda}\]

Answer 19

Now, substitute it back on the equation for N.
\[N(t) = N_0 e^{- \frac{ln(2) t}{t_{1/2}}}\]

Answer 20

Now, with this equation, you can calculate the mass of the element for every time t.

Answer 21

in order to make easier to make the table, set \(t = k*t_{1/2}\), where k = 0, 1, 2, ...