Related rates

[Chavi]

Does anybody have the formula for the surface area of a sphere?
Thanks

[superflya]

isnt it just

[the.watchman]

Yep, SA of a sphere

[samuch]

google is your best friend

http://en.wikipedia.org/wiki/Sphere

surface area= 4 x pi x r squared

[m@tty]

Just differentiate the volume of a sphere

...

[the.watchman]

Just differentiate the volume of a sphere

...

Wow, I never heard of that way before!
*makes mental note*

[m@tty]

Same with a circle. .

[Martoman]

Just differentiate the volume of a sphere

...

Wow, I never heard of that way before!
*makes mental note*

This is how i've always understood it, as its a small increase of the volume gets a infinitesimally small "film" around the sphere which in turn is its surface area. This, as m@tty says, can be seen in the circle as well.

[Chavi]

I have a few questions, if any1 could clarify:

1) A spherical ball of radius 4cm is coated with a layer of ice of uniform thickness. The ice melts at a uniform rate of 10cm^3 per hour. When the ice is 6cm thick, calculate the rate at which the surface area of the ice is decreasing.

2) A circular cylinder of height 6cm and base radius 4cm sits on a table with its axis vertical. A point source of light moves vertically upwards at a speed of 3cm/s above the central axis of the cylinder, casting a circular shadow on the table. Find the rate at which the radius of the shadow is decreasing when the light is at a distance 4cm above the top of the cylinder.

3) Assume that a raindrop is of spherical shape. If the raindrop accumulates moisture through condensation at a rate proportional to its surface area, show that the radius increases at a constant rate.

4) Grain is ejected from a chute at a rate of pi m^3/min and is forming a heap on a flat horizontal floor. The heap is in the form of a circular cone of semi vertical angle 45 degrees. Determine that rate at which the height is increasing 9min after opening the chute.

Thanks a million

[Martoman]

is the answer to the first question -2?

*edit* and q4, ? These could be horribly wrong tho 

[Yitzi_K]

I have a few questions, if any1 could clarify:

1) A spherical ball of radius 4cm is coated with a layer of ice of uniform thickness. The ice melts at a uniform rate of 10cm^3 per hour. When the ice is 6cm thick, calculate the rate at which the surface area of the ice is decreasing.

2) A circular cylinder of height 6cm and base radius 4cm sits on a table with its axis vertical. A point source of light moves vertically upwards at a speed of 3cm/s above the central axis of the cylinder, casting a circular shadow on the table. Find the rate at which the radius of the shadow is decreasing when the light is at a distance 4cm above the top of the cylinder.

3) Assume that a raindrop is of spherical shape. If the raindrop accumulates moisture through condensation at a rate proportional to its surface area, show that the radius increases at a constant rate.

4) Grain is ejected from a chute at a rate of pi m^3/min and is forming a heap on a flat horizontal floor. The heap is in the form of a circular cone of semi vertical angle 45 degrees. Determine that rate at which the height is increasing 9min after opening the chute.

Thanks a million

is the answer to the first question -2?

*edit* and q4, ? These could be horribly wrong tho  

Yeh I got -2 for the first one also:

You're looking for , which is .

To find , we use the relation .

We're given as 10, and as is obviously , that gives .

Now we can find , as it's , which is . Since (as it's the radius of the ball plus the thickness of the ice) this simplifies to give us 2, and since it's decreasing, it becomes -2.


For question 2, I remember we did the exact same question in class. If I recall correctly, you have to use similar triangles to find some of the lengths which give you the related rates you need. It's a bloody hard question though 

[Martoman]

I looked at 2 and just thought "screw that" and moved on. If I get some time tonight i'll do it, but i doubt I will

[TrueTears]

Just differentiate the volume of a sphere

...

Wow, I never heard of that way before!
*makes mental note*

This is how i've always understood it, as its a small increase of the volume gets a infinitesimally small "film" around the sphere which in turn is its surface area. This, as m@tty says, can be seen in the circle as well.

yup all comes from those sexy riemann sums <3

even when u change to polar or spherical coordinates, they never fail