I'm not 100% sure how to go about these 3 questions (especially the related rates problem), so if you can help me and show how you did it, that would be great!!!
If you scratch my back, I will scratch your back. Lmk

2 Half-life Problems:
Cobalt-60 has a half life of 5.24 years.
a.) Determine an exponential model for y, the mass of Cobalt-60 remaining at time t (measured in years), if the initial mass is 80 mg.





b.) How long will it take for the mass to decay to 3 mg?




1 Related Rates Problem:
A waterskier skies over the ramp at a speed of 30 feet per second. The ramp is 4 ft tall and 15ft long (the hypotenuse of this triangle is not given). How fast is she rising as she leaves the ramp?

1 Related Rates Problem:
A waterskier skies over the ramp at a speed of 30 feet per second. The ramp is 4 ft tall and 15ft long (the hypotenuse of this triangle is not given). How fast is she rising as she leaves the ramp?

first you need to find the hypotenuse, so use pythagorean thereom:

a^2 + b^2 = c^2
c = sqrt(15^2 + 4^2) = sqrt(41) ~ 15.5 ft

At 30 ft/s. this takes sqrt(241) / 30 s.

During that time she rises 4ft.

Rate of rise is therefore:
4 * 30 / sqrt(241)
= 7.73 ft/s.

For the related rates problem I also found this:

let ds/dt be the skier's speed - 30 ft/s
and dh/dt be her change in height with respect to time

dh/dt = 4/(241^0.5) * ds/dt
dh/dt = 7.73 ft/s


What I'm not sure of is where this 241 comes from in both answers.

After I know where this 241 comes from, I should be set on this Problem

This post was edited by CamelFinger on Nov 2 2013 02:19pm

it was a typo.

c = sqrt(15^2 + 4^2) = sqrt(241) ~ 15.5 ft

we can use the equation

y = A (1 - R) ^x

A is your initial amount
R is your Rate of Decay in this case (hence the minus sign). Often will be expressed in terms of PERCENTS
and X is the number of TIME intervals that have passed

you may have seen eqns similar to this one in past classes like A = Pe^RT or A = A0 (1 + (r/n))^t


80mg is 8 x 10^-5 kg

so for A would just be y = 8 X 10^-5 (1 - 0.5)^t

we use 0.5 here as R since they told you it's a half life. half life by definition is the amount of time it takes for there to be HALF of your original sample. 100g how long to 50g -> 25g -> 12.5g -> etc.

Part b asks you for how long it will take to get there. plug in 3 mg for your final amount

3 = 8 (1 - 0.5)^t

3/8 = (1 -0.5)^t

in order to "bring the t down" we use ln

ln(3/8) = t ln(0.5)
t = 1.41 years

241 is your x^2 value, or the hypotenuse of the ramp

the main eqn would be

A = A0 ( 1 +/- R/N)^t

This covers both exponential growth and decay problems as well as bank interest ones. N is the amount of times its compounded. if you memorize...shoudl be pretty good

wouldnt y = A (1 - R) ^x give how much mass has decayed, instead of giving how much mass is remaining?

Is this equation correct? I am not really sure if it is an exponential model for mass remaining at time t

e

10char

This post was edited by nlin on Nov 2 2013 03:49pm

x is the number of time intervals passed

in this example it has to be x/5.24 cause of half lifes

sorry, haven't done math in ~4-5 years, pretty rusty.

new eqn shoudl be

y = A (1 - R) ^ t/half life

3 = 80(1-0.5) ^ t /5.24
3/80 = 0.5 ^ t/5.24
[ln(3/80)] = t/5.24 ln(0.5)
[ln(3/80)]/[ln(0.5)] = t/5.24

5.24[ln(3/80)]/[ln(0.5)] = t
t = 24.82 years

fas;oeikljf;aoiejf;woailjekf;owaiejf

This post was edited by nlin on Nov 2 2013 03:58pm