I need Calculus help!!!
a conical paper cup is 15 cm high and has a radius of 5 cm the water level is rising at 4 cm/s. how quickly is the water being poured into the cup when the water is 8 cm deep?
in pi form with cm^3/s.
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Let h represent the height of the water in the cup, r the radius of the circular top of
the water level, and V the volume of water in the cup.
The ratio of radius over height is constant for cones with the same angle, so because
the paper cup has radius 3 cm and height 10 cm, the ratio r/h is equal to 3 cm over
10 cm, or r/h = 3/10. This gives us that r = 3/10 · h.
Now we can relate the volume of water in the cup to the height and radius by the
equation for the volume of a cone, V = (1/3)πr2h. In particular, using our equation
for r in terms of h, this gives us
V(t)=(1/3)pi*· (3/10 · h(t))^2*h(t)=(3pi/100)*h(t)^3
Using implicit differentiation on this relation gives us V(t)=(9pi/100)*h(t)^2h'(t),
When h(t) = 5 cm, the rate of change of volume is of course V
(t) = 2 cm3/s (since
this value is constant), so substituting these values into the above relation tells us that
the water level is rising at a rate of h
(t) = 8/(9π) cm/s.
listen this is the solution of a sum i had ...but both these sums are same only the numbers are different
you put your numbers in the correct place and you'll get your answer
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A paper cup has the shape of a cone with height 10 cm and radius 3 cm (at
the top). If water is poured into the cup at a rate of 2 cm3/s, how fast is the water
level rising when the water is 5 cm deep?
this is the question which i have solved...probably it will help you