hi, can someone please help me with these questions please? thanks! 5) 6) A sphe
ID: 2833798 • Letter: H
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hi, can someone please help me with these questions please? thanks!
5)
6)
A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.2 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 10 cm. (Note the answer is a positive number). At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 24 knots and ship B is sailing north at 15 knots. How fast (in knots) is the distance between the ships changing at 7 PM? Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 12 feet high? Recall that the volume of a right circular cone with height h and radius of the base r is given by V = 1/3 pir2h The linearization at a = 0 to 3 + 4x is A + Bx where A is: and where D isExplanation / Answer
Volume of sphere = 4/3 pi r^3
use implicit differentiation
dv/dt = 4/3 pi (3r^2) dr/dt
diameter is changing at a rate of .2cm/min so radius is changing at .1cm/min = dr/dt
when diameter = 10 or radius = 5
dv/dt = 4/3 pi * 3 * 5^2 * .1 = 10pi cm^3/min
6. if x = horizontal distance between a and b and y is vertical distance between a and b and z is the hypotenuse or straight distance than
z^2 = x^2 +y^2 take the derivative
2z dz/dt = 2x dx/dt + 2y dy/dt divide by 2 and z
dz/dt = (x dx/dt + y dy/dt)/z
x = 50 + 7hrs(24) = 218
y = 15*7 = 105
z = sqrt(218^2 +105^2) = 242
dz/dt = (218*24+105*15)/242 = 28.13 knots
diameter and height are equal so 2r = h or r = h/2
v = 1/3 pi (h/2)^2 h = 1/3 pi h^3/4
dv/dt = pi h^2/4 dh/dt
dv/dt = 10 and h = 12
10/( pi12^2/4) = dh/dt = 5/18pi = .088 ft/min
linearization is basically the equation of the tangent line at the point given.
if x = 0 then y = sqrt 3
take the derivative of the function to find the slope
2 /sqrt(3+4x) so slope at a=0 is 2/sqrt 3
y - sqrt(3) = 2/sqrt(3) (x-0)
linearization is y = 2x/sqrt(3) +sqrt 3
A = sqrt 3 and B = 2/sqrt(3) or 2sqrt(3)/3
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