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Violagirl

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I was having a hard time trying to figure out how to solve these two problems. Any help I would greatly appreciate! Thanks! :)

1. A storm at sea has damaged an oil rig. Oil spills from the rupture at the constant rate of 60 ft[sup:1642yqrg]3[/sup:1642yqrg]/min, forming a slick that is circular in shape and 3 inches thick. How fast is the radius of the slick increasing when the radius is 70 ft?

2. A baseball diamond is a square with side 90 ft. A batter hits the ball and runs towards first base with a speed of 24 ft/sec. At what rate is her distance from second base decresing when she is halfway to first base?
 
let y be the distance runner is from 1st base
D is distance of runner to second base
dy/dt= 24 ft /sec

when y=45 ft what is dD/dt

D= [ (90-y)^2+90^2]^1/2 take derivative with respect to t
dD/dt=1/2 [ (90-y)^2+90^2]^-1/2 [ 2(90-y) (-dy/dt)
dD/dt= - (90-y) dy/dt /[ (90-y)^2+90^2]^1/2
dD/dt =-(45)(24)/[45^2+90^2]^1/2
dD/dt= -1080 /1025^1/2
dD/dt=-10.73312629 ft/sec answer

Arthur

pblm 1 to follow
 
\(\displaystyle 1) \ Volume \ of \ a \ right \ circular \ cylinder \ = \ \pi r^{2}h.\)

\(\displaystyle Given: \ h \ = \ \frac{1}{4}ft., \ a \ constant, \ \frac{dV}{dt} \ = \ \frac{60ft^{3}}{min}.\)

\(\displaystyle Find \ \frac{dr}{dt} \ when \ radius \ = \ 70ft.\)

\(\displaystyle \frac{dV}{dt} \ = \ 2\pi hr\frac{dr}{dt}, \ \frac{60ft^{3}}{min} \ = \ 2\pi(\frac{1}{4}ft)(70ft)\frac{dr}{dt}.\)

\(\displaystyle You \ should \ be \ able \ to \ take \ it \ from \ here.\)
 
If we pretend first base is at x=0 (orient the diamond so that the y axis is in line with first and second base), then the problem becomes easier:

D^2 = 90^2 + x^2

Take derivative implicitly:

2D*dD/dt =2x*dx/dt

Solve for Dd/dt

dD/dt = x/D*(dx/dt)

We know dx/dt = 24ft/sec, halfway betwwen 1st/2nd means x=-45, which means D^2=90^2+(-45)^2, so D~100.62:

dD/dt = (-45)/100.62*(24) ~ -10.73 ft/s

so its decreasing at a rate of 10.73 ft/s.


edited to fix spelling, minus sign
 
arthur ohlsten said:
let y be the distance runner is from 1st base
D is distance of runner to second base
dy/dt= 24 ft /sec

when y=45 ft what is dD/dt

D= [ (90-y)^2+90^2]^1/2 take derivative with respect to t
dD/dt=1/2 [ (90-y)^2+90^2]^-1/2 [ 2(90-y) (-dy/dt)
dD/dt= - (90-y) dy/dt /[ (90-y)^2+90^2]^1/2
dD/dt =-(45)(24)/[45^2+90^2]^1/2
dD/dt= -1080 /1025^1/2
dD/dt=-10.73312629 ft/sec answer

Arthur

pblm 1 to follow
Click to expand...

Solution above has WRONG assumptions.
 
Thanks for the help everyone!!

For the first problem, I did it twice but I may have gotten a calculating error. Is the answer around .022? :?
 
\(\displaystyle 2) \ Given: \ \frac{dx}{dt} \ = \ \frac{24ft}{sec}, \ y \ =90ft, \ a \ constant\)

\(\displaystyle Find \frac{dh}{dt} \ when \ x \ = \ 45ft.\)

\(\displaystyle h^{2} \ = \ x^{2}+y^{2} \ = \ 45^{2}+90^{2}, \ h \ = \ 45\sqrt5.\)

\(\displaystyle Hence, \ 2h\frac{dh}{dt} \ = \ 2x\frac{dx}{dt} +0, \ h\frac{dh}{dt} \ = \ x\frac{dx}{dt},\)

\(\displaystyle = \ 45\sqrt5 \frac{dh}{dt} \ = \ (45)(24), \ \frac{dh}{dt} \ = \ \frac{(45)(24)}{45\sqrt5} \ = \ 10.733ft/sec\)
 
Thanks for clarifying that BigGlenntheheavy!

Did I get the correct answer on the first problem? I got .022 as an answer.
 
BigGlenntheHeavy said:
\(\displaystyle 2) \ Given: \ \frac{dx}{dt} \ = \ \frac{24ft}{sec}, \ y \ =90ft, \ a \ constant\)

\(\displaystyle Find \frac{dh}{dt} \ when \ x \ = \ 45ft.\)

\(\displaystyle h^{2} \ = \ x^{2}+y^{2} \ = \ 45^{2}+90^{2}, \ h \ = \ 45\sqrt5.\)

\(\displaystyle Hence, \ 2h\frac{dh}{dt} \ = \ 2x\frac{dx}{dt} +0, \ h\frac{dh}{dt} \ = \ x\frac{dx}{dt},\)

\(\displaystyle = \ 45\sqrt5 \frac{dh}{dt} \ = \ (45)(24), \ \frac{dh}{dt} \ = \ \frac{(45)(24)}{45\sqrt5} \ = \ 10.733ft/sec\)
Click to expand...

The distance between the runner and the second base is decreasing ? hence 'h' is decreasing ? hence dh/dt should be 'negative' ? hence the solution above is based on wrong assumption.
 
BigGlenntheHeavy said:
Are we splitting hairs, Subhotosh Khan?
Click to expand...

No - just being correct.

If the "assumption" in your problem is incorrect - then the "mechanics" does not matter. As Galactus would say - "technology can do the mechanics".
 
\(\displaystyle \frac{dh}{dt} \ = \ 10.733 ft/sec, hence \ it \ is \ decreasing \ 10.733ft/sec\)

\(\displaystyle If \ it \ were \ decreasing \ -10.733ft/sec, \ it \ would \ be \ increasing. \ (-)(-) \ = \ (+)\)
 
Glen dh/dt is negative (and continuous), hence it is decreasing. See my post above.
 
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