I posted this question a few days ago and someone answered it but I just wanted to make sure that I copied down the correct answers and didn't misunderstand anything.
Related Rates:
Gas is escaping a spherical balloon at the rate of 2ft^3/min. How fast is the surface area shrinking (ds/dt) when the radius is 12ft? ( A sphere of raius r has volume= 4/3 pir^3 and surface area S=4pir^2.)
Remember that ds/dt=ds/dr X dr/dt
Step 1: Find ds/dr
S=4pir^2 so ds/dr= (4pi)(2r) = 8pi r
Step 2: Fid dr/dt (Hint: dv/dt=dv/dr X dr/dt)
V=4/3pir^3 dv/dr=4pir^2 2=(4pir^2)(dr/dt) dr/dt=1/2pir^2
Step 3: Find ds/dt
ds/dt= (8pir)(1/2pir^2) = 4/r
Step 4: Evaluate ds/dt when the radius is 12ft.
r=12 (8)(3.14)(12) X (1/2(3.14)(12)^2) =4/12
(301.44)(1/904.32)=4/12
Are these correct?
correct, except reduce 4/12 to 1/3
I might have done it this way
V = (4/3)πr^3
dV/dt = 4πr^2 dr/dt , but when r=12, dV/dt = 2
2 = 4π(144) dr/dt
dr/dt = 1/(288π)
A = 4πr^2
dA/dt = 8πr dr/dt
= 8π(12)(1/(288π) = 1/3
Ok thank you!
welcome
Let's go through the steps again and check the answers:
Step 1: Finding ds/dr
The surface area is given by S = 4πr^2. Taking the derivative of S with respect to r, we get dS/dr = 8πr.
Your answer of ds/dr = 8πr is correct.
Step 2: Finding dr/dt
The volume of the sphere is given by V = (4/3)πr^3. Taking the derivative of V with respect to r, we get dV/dr = 4πr^2. Given that dV/dt = 2 ft^3/min, we can solve for dr/dt:
2 = (4πr^2)(dr/dt)
dr/dt = 2 / (4πr^2)
dr/dt = 1 / (2πr^2)
Your answer of dr/dt = 1 / (2πr^2) is correct.
Step 3: Finding ds/dt
Using the chain rule, we can calculate ds/dt as ds/dt = (ds/dr) ⋅ (dr/dt).
Substituting the values we found in step 1 and step 2:
ds/dt = (8πr) ⋅ (1 / (2πr^2))
ds/dt = 4/r
Your answer of ds/dt = 4/r is correct.
Step 4: Evaluating ds/dt for r = 12 ft
Substituting r = 12 ft into the expression we derived for ds/dt:
ds/dt = 4/12
ds/dt = 1/3 ft^2/min
So, the correct answer is ds/dt = 1/3 ft^2/min.
Therefore, based on the calculations, it appears that your original answers are correct.