Sand is poured at the rate of 10 m cube /min so as to form a conical pile whose altitude is always equal to the radius of its base
a) How fast is the radius increasing when it is 5 meters?
b) Find the rate at which the area of its base is increasing when the radius is 5m
no ....
dV/dt = π r^2 dr/dt
10 = π(25)dr/dt
dr/dt = 10/(25π) = .127...
which of course means your second answer is also incorrect
dA/dt = 2πr dr/dt
= 2π(5)(.127...) = 4
You also had 4, but should not have obtained that answer if you used
the wrong answer to a)
very strange
So in V = (1/3)π r^2 h, the value of h = r, so we get
V = (1/3)π r^3
dV/dt = π r^2 dr/dt
a) plug in dV/dt = 10 m^3 and r = 5m, then solve for dr/dt
b) Area of base = πr^2
dA/dt = 2πr dr/dt
you have dr/dt from a) and r = 5
a) Well, if the altitude is always equal to the radius, we can call both of them 'r'. Now, to find how fast the radius is increasing, we need to use the volume formula of a cone. The volume (V) of a cone is given by V = (1/3)πr^2h, where r is the radius and h is the height. In this case, h = r.
Since the sand is poured at a rate of 10 m^3/min, we have dV/dt = 10 m^3/min. Now, we can differentiate the volume formula with respect to t (time) to find how the radius is changing with time:
dV/dt = (1/3)π(2r(dr/dt) + r^2)
Simplifying a bit, we get 10 = (2/3)πr(dr/dt + r)
Now, we can solve for dr/dt by plugging in r = 5 (since we want to find the rate when the radius is 5 meters):
10 = (2/3)π(5)(dr/dt + 5)
10 = (10/3)π(dr/dt + 5)
1 = (1/3)π(dr/dt + 5)
(1/π) - 5/π = dr/dt
dr/dt = (1 - 5π)/π
So, the rate at which the radius is increasing when it is 5 meters is approximately equal to (1 - 5π)/π.
b) To find the rate at which the area of the base is increasing, we need to differentiate the formula for the area (A) of a circle with respect to t:
A = πr^2
dA/dt = 2πr(dr/dt)
Now, we know that dr/dt = (1 - 5π)/π, so we can substitute that into the equation:
dA/dt = 2π(5)(1 - 5π)/π
dA/dt = 10(1 - 5π)
So, the rate at which the area of the base is increasing when the radius is 5 meters is approximately equal to 10 - 50π. Just don't be square with your calculations!
To answer these questions, we can use the formulas for the volume and surface area of a cone. The volume of a cone is given by V = (1/3)πr²h, where V is the volume, r is the radius, and h is the height (altitude). The surface area of a cone is given by A = πr(r+√(r²+h²)), where A is the surface area.
a) To find how fast the radius is increasing (dr/dt) when it is 5 meters, we need to differentiate the volume equation with respect to time (t). Let's denote r as a function of time, r(t). Differentiating both sides of the volume equation gives dV/dt = (1/3)π(2rr')h + (1/3)πr²h', where r' = dr/dt (the rate at which the radius is changing), and h' = dh/dt (the rate at which the height is changing). Since the height is always equal to the radius, we have h' = r'.
Given that the sand is being poured at the rate of 10 m³/min, we know that dh/dt = 10. Plugging in all the values into the equation, we get dV/dt = (1/3)π(2rr')(r+r'). Rearranging the equation, we have r' = (3dV/dt)/(2πr²+πr³).
Now, substituting r = 5 and dV/dt = 10 into the equation, we can calculate the rate at which the radius is increasing when it is 5 meters.
b) To find the rate at which the area of the base is increasing (dA/dt) when the radius is 5 meters, we need to differentiate the surface area equation with respect to time (t). Let's denote A as a function of time, A(t). Differentiating both sides of the surface area equation gives dA/dt = π(2rr') + π(r+√(r²+h²))(r'), where r' = dr/dt (the rate at which the radius is changing), and h' = r' (since the height is always equal to the radius).
Now, substituting r = 5 and r' (which we found in part a) into the equation, we can calculate the rate at which the area of the base is increasing when the radius is 5 meters.
so sir reiny the answer i got in your solution is
a.0.4 m/min
b.4 m/min
that is right?