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jsmith613
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Homework Statement
I have attached a link along with my working
please can someone help me
http://s359.photobucket.com/albums/oo40/jsmith613/?action=view¤t=Math.png
jsmith613 said:Homework Statement
I have attached a link along with my working
please can someone help me
http://s359.photobucket.com/albums/oo40/jsmith613/?action=view¤t=Math.png
Homework Equations
The Attempt at a Solution
dv/dt = 4 cm3 min-1Mark44 said:Please post the problem and your work here. IMO, it's a pain in the butt to have to open a web page to see the problem and the work, plus I can't insert a comment at the appropriate place where there's an error.
You're ignoring the relationship between r and h.jsmith613 said:dv/dt = 4 cm3 min-1
tan(60) = r/h
r = √3* h
V = (1/3) π r2h
dv/dh = (1/3) π r2
jsmith613 said:dh/dt = (dh/dv) * (dv/dt)
= 1/(1/3) π r2 * 4
= 12/(pi r2)
for h = 4
dh/dt = 0.079577
The answer given is 0.0265 cm/min. why?
I don't know how do use latex - see post for a clearer solution if you get lost!
The formula for calculating the rate of change for cones is (change in height)/(change in radius).
To find the change in height, subtract the initial height from the final height. To find the change in radius, subtract the initial radius from the final radius.
Yes, you can use any units for the height and radius as long as they are consistent. For example, if you use meters for height, you should also use meters for radius.
The unit for the rate of change for cones is (unit of height)/(unit of radius), such as meters/centimeters or inches/feet.
No, the rate of change for cones only calculates the change in height over the change in radius. To find the volume or surface area, you will need to use the formulas specific to those measurements.