If the radius of the sphere is increasing at a rate of 2 inches per second, how fast , in cubic inches per ...

Question

Solve and show the work for this problem:
This is a calculus problem -

If the radius of the sphere is increasing at a rate of 2 inches per second, how fast , in cubic centimeter inches per second is the volume increasing when the radius is 10 inches?

Thanks

Answer 1

vol = 4 pi r^3/3
vol ' = 4 pi r^2 r'
= 4 pi 100 * 2
~ 2513.274 cubic in / sec

Answer 2

V=4/3 *pi*r^3 dV/dt= dV/dr*dr/dt = 4pir^2 *dr/dt = 4*pi*100*2= 800 pi cubic inch/s

Answer 3

The radius of a sphere is increasing at a rate of 2 inches per second, which means

dr/dt = 2

We want to know how fast the volume is increasing; i.e. we want to know

dV/dt = ?

What we want to do is relate our unknowns V and r. They are related by the following equation, which is the volume of a sphere.

V = (4/3)pi(r^3)

At this point, we want to differentiate with respect to t. Doing so (and remember to ignore the constants (4/3)pi when differentiating), we get

dV/dt = (4/3)pi(3r^2)(dr/dt)
dV/dt = 4pi(r^2)(dr/dt)

Here is where we plug in our known rates. dr/dt is given to be 2, so

dV/dt = 4pi(r^2)(2)
dV/dt = 8pi(r^2)

Here is where we insert our "when" statement. Every related rates question has a "when" statement.

*When the radius is 10 inches..."
This translates to r = 10, so it is at this point we plug in r = 10.

dV/dt = 8pi(10)^2
dV/dt = 8pi(100)
dV/dt = 800pi

Now, we make our concluding statement.

"The sphere is increasing at a rate of 800pi cubic inches per second."

Answer 4

first
volume of sphere = 4/3 pi * r^3 ( i will refer to pi by just p )
let radius changes from 10 to 10 + h
r1 = 10 ----------------------- v1 = 4/3 * p*10^3 = 4000/3 p
r2 = 10+h ---------------------v2 = 4/3 * p*(10+h)^3 =
4/3p( 1000+300h+30h^2 +h^3)
v2=4000/3p + 400ph + 40ph^2 + 4/3p. h^3

change in v
= v2-v1=4000/3p + 400ph + 40ph^2 + 4/3p. h^3 - 4000/3 p
delta v = 400ph + 40ph^2 + 4/3p. h^3
the average rate of change = delta v / delta r
= (400ph + 40ph^2 + 4/3p. h^3 )/h
average = 400p + 40ph + 4/3 p h^2
the rate of change = limit of average rate at h tends to 0
then it is equal to = 400p + 40p*0 + 4/3 *p*(0)^2
the rate = 400 p
i hope that helppppssss

Answer 5

V= 4/3pi r^3
dV/dt = dV/dr* dr/dt
dV/dr= 4pir^2 and dr/dt= 2inches/second
so dV/dt= 4pi*100(2) =2513.27 cubic inches/s
1 cubic inch =16.39 cm^3

Answer 6

dR/dt=2

V=4/3*pi*R³
dV/dR = 4*pi*R²

Change rule:

dV/dt
=
dV/dR * dR/dt
=
4*pi*R² * 2
=
8*pi*R²

So plug R=10 into that, and there's your answer.