can someone check these for me:
18. All edges of a cube are expanding at a rate of 6 cm per second. How fast is the volume changing when each edge is a) 2 cm and b) 10 cm.
a)72 cm^3/sec b)1800 cm^3/sec
20. The formula for the volume of a cone is v=(1/3)(pi)(r^2)h. Find the rates of change of the volume if dr/dt is 2 in per min and h=3r when a) r=6 in and b)r=24 in
a)216(pi) in^2/min
b)12864(pi) in^2/sec
thanks
18. correct
20 a) correct
b) I got 3456π inches^2/min (why have you got in^2/sec ? )
To solve the given problems, we'll need to use calculus and apply the chain rule to find the rates of change.
Question 18:
a) To find how fast the volume of the cube is changing when each edge is 2 cm, we need to differentiate the formula for the volume of a cube, V = s^3, where s is the length of each edge. Taking the derivative with respect to time t, we have:
dV/dt = d/dt (s^3)
To express this in terms of rate of change, we need to find ds/dt, which represents the rate at which each edge is expanding. Given that ds/dt = 6 cm/sec, we have:
dV/dt = d/dt (2^3)
= 3(2^2)(ds/dt)
= 3(4)(6)
= 72 cm^3/sec
Therefore, when each edge of the cube is 2 cm, the volume is changing at a rate of 72 cm^3/sec.
b) Similarly, when each edge is 10 cm, the rate of change of the volume can be found as follows:
dV/dt = d/dt (10^3)
= 3(10^2)(ds/dt)
= 3(100)(6)
= 1800 cm^3/sec
Therefore, when each edge of the cube is 10 cm, the volume is changing at a rate of 1800 cm^3/sec.
Question 20:
a) To find the rate of change of the volume of a cone when r = 6 in, given that dr/dt = 2 in/min and h = 3r, we can substitute the given values into the formula for the volume of a cone:
V = (1/3)(π)(r^2)h
Substituting h = 3r, we get:
V = (1/3)(π)(r^2)(3r)
= (π/3)(3r^3)
= (πr^3)
Now, take the derivative of V with respect to t:
dV/dt = d/dt (πr^3)
Applying the chain rule, we have:
dV/dt = 3(πr^2)(dr/dt)
Substituting the given values, we have:
dV/dt = 3(π)(6^2)(2)
= 3(π)(36)(2)
= 216π in^2/min
Therefore, when r = 6 in, the volume of the cone is changing at a rate of 216π in^2/min.
b) Following the same process, when r = 24 in, we have:
dV/dt = 3(π)(24^2)(2)
= 3(π)(576)(2)
= 3,456π in^2/min
Therefore, when r = 24 in, the volume of the cone is changing at a rate of 3,456π in^2/min.
Make sure to double-check the calculations and units as you work through these problems.