LoveToStack
18-06-2011, 12:28 PM
Hey, just doing some revision and need to know whether or not I have the right answer here. I've got one answer, the book says another.
Question:
A hollow cone, with a semi-vertical angle of 60°, is held vertex downwards with it's axis vertical. Water drips into the cone at a constant rate of 4cm3 per minute. Find the rate at which the depth of the water is increasing when the water is 4cm deep.
If that description is unclear, just picture a conical water cup with the point facing downward, that's the jist of it. Also the angle made between the slope of the cone and the exact centre is 60°.
Here's my working:
Since we're given the angle 60° and the depth of the water as 4cm, we can work out the radius of the cross section of the cone (or the top of the water) at that depth using pythagoras:
Tan60 = radius / 4
radius = 4√3
Now we know the radius we can use related rates of change to work out dh/dt, where h is the depth of the water.
We are given dV/dt in the question as 4cm3.
And we can use the chain rule to say:
dV/dt = (dh/dt) x (dV/dh)
So now I just needed to work out dV/dh and I could solve for dh/dt:
Volume of a cone = (1/3)(pi)(r^2)h
But I know at this point that the radius = 4√3 so:
Volume of a cone = (1/3)(pi)(4√3)^2 x h = ((48pi)/3)h
So dV/dh = (48pi)/3
It follows then that:
dV/dt = (dh/dt) x (dV/dh)
4 = (dh/dt) x (48pi)/3
1/(4pi) = dh/dt
Final answer: dh/dt = 0.0796cm / minute
But the book gets a different answer. For reference, the book's answer is 0.0265cm / minute.
I just need a second opinion because I have my exam on Monday and I'm a bit sketchy on related rates of change (among other things lol...). Thanks.
Question:
A hollow cone, with a semi-vertical angle of 60°, is held vertex downwards with it's axis vertical. Water drips into the cone at a constant rate of 4cm3 per minute. Find the rate at which the depth of the water is increasing when the water is 4cm deep.
If that description is unclear, just picture a conical water cup with the point facing downward, that's the jist of it. Also the angle made between the slope of the cone and the exact centre is 60°.
Here's my working:
Since we're given the angle 60° and the depth of the water as 4cm, we can work out the radius of the cross section of the cone (or the top of the water) at that depth using pythagoras:
Tan60 = radius / 4
radius = 4√3
Now we know the radius we can use related rates of change to work out dh/dt, where h is the depth of the water.
We are given dV/dt in the question as 4cm3.
And we can use the chain rule to say:
dV/dt = (dh/dt) x (dV/dh)
So now I just needed to work out dV/dh and I could solve for dh/dt:
Volume of a cone = (1/3)(pi)(r^2)h
But I know at this point that the radius = 4√3 so:
Volume of a cone = (1/3)(pi)(4√3)^2 x h = ((48pi)/3)h
So dV/dh = (48pi)/3
It follows then that:
dV/dt = (dh/dt) x (dV/dh)
4 = (dh/dt) x (48pi)/3
1/(4pi) = dh/dt
Final answer: dh/dt = 0.0796cm / minute
But the book gets a different answer. For reference, the book's answer is 0.0265cm / minute.
I just need a second opinion because I have my exam on Monday and I'm a bit sketchy on related rates of change (among other things lol...). Thanks.