The volume V of a right circular cone of radius r and height h is given by V = 1 3 π r 2 h . Suppose that the height decreases from 20 in to 19.95 in and the radius increases from 4 in to 4.05 in. Compare the change in volume of the cone with an approximation of this change using a total differential.
The volume V of a right circular cone of radius r and height h is given by V = 1 3 π r 2 h . Suppose that the height decreases from 20 in to 19.95 in and the radius increases from 4 in to 4.05 in. Compare the change in volume of the cone with an approximation of this change using a total differential.
The volume V of a right circular cone of radius r and height h is given by
V
=
1
3
π
r
2
h
.
Suppose that the height decreases from 20 in to 19.95 in and the radius increases from 4 in to 4.05 in. Compare the change in volume of the cone with an approximation of this change using a total differential.
With integration, one of the major concepts of calculus. Differentiation is the derivative or rate of change of a function with respect to the independent variable.
An outdoor decorative pond in the shape of a hemispherical tank is to be filled with water pumped into the tank through an inlet in its bottom. Suppose that the radius of the tank is R = 10 ft, that water is pumped in at a rate ofT ft/min,
and that the tank is initially empty. As the tank fills, it loses water through evaporation. Assume that the rate of evaporation is proportional to the area A of the surface of the water and that the constant of proportionality is k = 0.01.
Output: water evaporates
at rate proportional
to area A of surface
ER-
Input: water pumped in
at rate 7 ft/min
(a) hemispherical tank
(b) cross-section of tank
(a) The rate of change
dv
of the volume of the water at time t is a net rate. Use this net rate to determine a differential equation for the height h of the water at time t. The volume of the water shown in the figure is V = TRh -Th,
dt
where R = 10. Express the area of the surface of the water A = Tr2 in terms of h.
dh
dt
(b) Solve the differential…
the weight W of a steel ball bearing varies directly with the cube of the bearing's radius r according to the formula W= 4/3 pi(p)(r)^3, where p is the density of the steel. The surface area of a bearing varies directly as the square of its radius because A = 4 pi(r^2)
a. Express the weight W of a bearing in terms of its surface area
b. Express the bearing's surface area A in terms of its weight.
c. For steel, p = 7.85 g/cm^3. What s the surface area of a bearing weighing 0.62 g?
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