Expert: Alon Mandes - 11/3/2009

Question
Hey there I have a test tomorrow nigh for Calc and I just can't seem to get the answers for these questions, i've been trying hard, but always come out wrong. Please reply by tomorrow any time if you can it will be GREATLY appreciated. PLEASE IF YOU CAN"T GET ALL SHOW ME HOW TO DO ANY THAT YOU CAN DO. IF YOU THINK ITS TOO MUCH PLEASE DO WHAT YOU CAN.


1.A baseball diamond is a square with side 90 ft. A batte1r hits the ball and runs toward first base with a speed of 22 ft/s.

a)At what rate is his distance from second base decreasing when he is halfway to first base?

b)At what rate is his distance from third base increasing at the same moment?

2.Whats the 30th derivative of cos(2x)?

3. An object with weight W is dragged along a horizontal plane by a force acting along a rope attached to the object. If the rope makes an angle x with the plane, then the magnitude of the force is given by the following equation, where μ is a constant called the coefficient of friction.

μW / μsin(x)+cos(x)

a.Find the rate of change of F with respect to x.

b.When is this rate of change equal to 0?

c.If W = 60 lbs and μ = 0.8, draw the graph of F as a function of x and use it to locate the value of x for which dF / dx = 0. (Round the answer to two decimal places.)


4.Each side of a square is increasing at a rate of 8 cm/s. At what rate is the area of the square increasing when the area of the square is 49 cm^2?  (in cm^2/s)


5.Scientist can determine the age of ancient objects by a method called radiocarbon dating. The bombardment of the upper atmosphere by cosmic rays converts nitrogen to a radioactive isotope of carbon, 14C, with a half-life of about 5730 years. Vegetation absorbs carbon dioxide through the atmosphere and animal life assimilates 14C through food chains. When a plant or animal dies, it stops replacing its carbon and the amount of 14C begins to decrease through radioactive decay. Therefore, the level of radioactivity must also decay exponentially. A parchment fragment was discovered that had about 74% as much 14C radioactivity as does plant material on Earth today. Estimate the age of the parchment in years.

6.If h(x) is given below, where f(3) = 7 and f '(3) = 5, find h'(3)

h(x)=sqrt(7+6f(x))

7.Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the equation PV = C, where C is a constant. Suppose that at a certain instant the volume is 300 cm3, the pressure is 150 kPa, and the pressure is increasing at a rate of 40 kPa/min. At what rate is the volume decreasing at this instant? cm^3/min

THANKS SO MUCH! George

Answer
#Q1
a)Let's set x(t) the distance from the 2nd base . The distance from the 1st base will be
90-22t . According to Pythagoras : x(t)=√[90²+(90-22t)²]=√[8100+8100-484t+t²]=√[16200-484t+t²].
Therefore: x'(t)=(-484+2t)/( 2√[16200-484t+t²] ). When he is midway , 22to=45 -->to=45/22 .
So to find x'(45/22) We simply substitute t=45/22 into x'(t) .
b)Let's set y(t) the distance from the 2nd base . The distance from the 1st base will be
90-22t . According to Pythagoras : y(t)=√[90²+(22t)²]=√[8100+484t²].
Therefore: y'(t)=968t/( 2√[8100+484t²] ). When he is midway , 22to=45 -->to=45/22 .
So to find y'(45/22) We simply substitute t=45/22 into y'(t) .

#Q2
cos(2x)'=-2sin(2x)
cos(2x)''=4cos(2x)
cos(2x)'''=-8sin(2x)
cos(2x)''''=16cos(2x)
Therefore :
30th derivative of cos(2x) = 2^(30)cos(2x) .

#Q3  
a)F(x)=μW/[μsin(x)+cos(x)]. To find rate of change, we derive the function . I will use the rule:
[1/f]'=f'/f² . So,
F'(x)=μW[μcos(x)-sin(x)]/[μsin(x)+cos(x)]².
b)F'(x)=0 When μcos(x)-sin(x)=0 --> μcos(x)=sin(x) --> Tang(x)=μ --> x=Arctang[μ] .

#Q4
Area=Side²  -> dA/dt=2S(dS/dt) . when the area is 49 then the side will be 7 . Therefore :
dA/dt=2*7*8=112 cm/s .

Q#6
h(x)=√[7+6f(x)] . Then h'(x)=6f'(x)/( 2√[7+6f(x)] ). Therefore ,
h'(3)=6f'(3)/( 2√[7+6f(3)] )=6*7/( 2√[7+6*5] )=42/( 2√[37] )= 3.45

Q#7
Let's set pressure as p(t) , %26 volume as v(t) . We know that p(t)v(t)=C . Now let's derive
this equation : To derive it we use the product rule of derivation : [fg]'=f'g+fg'. Thus,
[p(t)v(t)]'=C' . C'=0 because its constant . Therefore :
p'(t)v(t)+p(t)v'(t)=0 . We need to find v'(t), given that p(t)=150,v(t)=300 %26 p'(t)=40 :
40*300+150*v'(t)=0 --> v'(t)= -40*300/150 = -80 cm^3/min .

Hope This will help. Good luck,
Alon.