SOLUTION: Given a right circular cone leaking water (original dimensions of r=6m h=8m
1. Rewrite the formula for V in terms of the radius only.
2. Rewrite the formula for V in terms of h
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-> SOLUTION: Given a right circular cone leaking water (original dimensions of r=6m h=8m
1. Rewrite the formula for V in terms of the radius only.
2. Rewrite the formula for V in terms of h
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Question 54765: Given a right circular cone leaking water (original dimensions of r=6m h=8m
1. Rewrite the formula for V in terms of the radius only.
2. Rewrite the formula for V in terms of height only.
3. If the volume is decreasing at a rate of 81/32 (pie) m^3/min. how fast is the height dropping when h=3m?
Please help...i've search so many places for help. Lillian
You can put this solution on YOUR website! Given a right circular cone leaking water (original dimensions of r=6m h=8m
1. Rewrite the formula for V in terms of the radius only.
WHERE IS THE FORMULA?OK LET ME ASSUME
V=PI*R^2*H/3..IF YOU WANT ONLY INTERMS OF RADIUS WE HAVE TO BE GIVEN SOME RELATION BETWEEN HEIGHT AND RADIUS..AGAIN LET ME ASSUME SLANT HEIGHT IS GIVEN AS 10.(THIS CORRESPONDS WITH H=8 FOR R=6,SINCE R^2+H^2=6^2+8^2=36+64=10^2=100).THEN
R^2+H^2=10^2
H= SQRT (100-R^2)
HENCE
V=PI*R^2/3
2. Rewrite the formula for V in terms of height only.
SAME COMMENTS AS ABOVE APPLY..MAKING SAME ASSUMPTIONS
R^2=100-H^2
V=PI(100-H^2)*H/3
3. If the volume is decreasing at a rate of 81/32 (pie) m^3/min. how fast is the height dropping when h=3m?
V=PI(100-H^2)H/3=(100H-H^3)PI/3
DV/DT = (100-3H^2)(PI/3)DH/DT
DV/DT = 81PI/32
H=3
DH/DT = {(81PI/32)}/{(100-3*3^2)PI/3}=(81*3)/(32*73)=243/2336=0.104
Please help...i've search so many places for help. Lillian
