Today: sections 4.1, 4.7
Example 4.1
A spherical balloon is being filled with air at the constant rate of 2 cm^3 /s,
means
the changing rate of the volume of this spherical balloon, is 2 cm^3/s
in terms of 
#card=math&code=V%28t%29&id=u4nve): 
%20%3D%202%20cm%5E3%2Fs#card=math&code=V%27%28t%29%20%3D%202%20cm%5E3%2Fs&id=f7afD)
(Note that both the volume and radius of the balloon are functions of time, 
)
From %20%3D%204%5Cpi%20%5Br(t)%5D%5E2%20%5Ccdot%20r’(t)#card=math&code=V%27%28t%29%20%3D%204%5Cpi%20%5Br%28t%29%5D%5E2%20%5Ccdot%20r%27%28t%29&id=MwtEY), we write with the units:
V’(t) — cm^3/s
4
— (none)
r(t)— cm, 
%5D%5E2#card=math&code=%5Br%28t%29%5D%5E2&id=yDDjj) — cm^2;
%20%3D%20%5Cfrac%7B%20d%20r(t)%7D%7Bdt%7D#card=math&code=r%27%28t%29%20%3D%20%5Cfrac%7B%20d%20r%28t%29%7D%7Bdt%7D&id=fVkFp)
2 cm^3/s = 4 
(
#card=math&code=r%28t%29&id=MqSW5) cm^2) 
#card=math&code=r%27%28t%29&id=La53j)
cm/s
dimension analysis
Example 4.2
(units: ft^2) doing computation with units,
A = the man, B = the air plane, C = the position on the ground directly under the air plane. 
%3D%20AB#card=math&code=s%20%3D%20s%28t%29%3D%20AB&id=NA29N).
.
Taking square roots on both sides, we get 
To compute 
%3DAB#card=math&code=s%27%3Ds%27%28t%29%3DAB&id=yvOwT) . There is a constant: 
.
%5E2%20%3D%20AC(t)%5E2%20%2B%204000%20ft%5E2#card=math&code=AB%28t%29%5E2%20%3D%20AC%28t%29%5E2%20%2B%204000%20ft%5E2&id=ZKdsz)
The speed of the air plane is 600 ft/sec. That is, 
%20%3D%20AC’(t)%20%20%3D%20600%20ft%2Fs#card=math&code=x%27%28t%29%20%3D%20AC%27%28t%29%20%20%3D%20600%20ft%2Fs&id=Ta6AG).
Example 4.32;
maximize 
%20%3D%20x(100-%202x)%3D%20100%20x%20-%202x%5E2#card=math&code=A%28x%29%20%3D%20x%28100-%202x%29%3D%20100%20x%20-%202x%5E2&id=lumgI).
. 
%20%3D%20100%20-%204x%3D4(25-x)#card=math&code=A%27%28x%29%20%3D%20100%20-%204x%3D4%2825-x%29&id=rOoot). 
is a critical point. Applying the first derivative test,
we see that 
is the point that #card=math&code=A%28x%29&id=uGGEF) attains its maximum.
Another way: the arithmetic-geometric mean inequality.
%20%3D%20100%20x%20-%202x%5E2%20%3D%202x(50-x)%20%5Cleq%202%20(%5Cfrac%7Bx%2B(50-x)%7D%7B2%7D)%5E2%20%3D%202%5Cfrac%7B50%5E2%7D%7B4%7D%20%3D%20%5Cfrac%7B2500%7D%7B2%7D%3D1250#card=math&code=A%28x%29%20%3D%20100%20x%20-%202x%5E2%20%3D%202x%2850-x%29%20%5Cleq%202%20%28%5Cfrac%7Bx%2B%2850-x%29%7D%7B2%7D%29%5E2%20%3D%202%5Cfrac%7B50%5E2%7D%7B4%7D%20%3D%20%5Cfrac%7B2500%7D%7B2%7D%3D1250&id=nMZIe).
because 
%3D50#card=math&code=x%20%2B%20%2850-x%29%3D50&id=d0pF7) is a constant.
For the reason of completion, compare the values: 
%3D0%2C%20A(50)%20%3D%200%2C%20A(25)%3D1250#card=math&code=A%280%29%3D0%2C%20A%2850%29%20%3D%200%2C%20A%2825%29%3D1250&id=J5xFN).
