<h3>Given</h3>
- a cone of height 0.4 m and diameter 0.3 m
- filling at the rate 0.004 m³/s
- fill height of 0.2 m at the time of interest
<h3>Find</h3>
- the rate of change of fill height at the time of interest
<h3>Solution</h3>
The cone is filled to half its depth at the time of interest, so the surface area of the filled portion will be (1/2)² times the surface area of the top of the cone. The filled portion has an area of
... A = (1/4)(π/4)d² = (π/16)(0.3 m)² = 0.09π/16 m²
This area multiplied by the rate of change of fill height (dh/dt) will give the rate of change of volume.
... (0.09π/16 m²)×dh/dt = dV/dt = 0.004 m³/s
Dividing by the coefficient of dh/dt, we get
... dh/dt = 0.004·16/(0.09π) m/s
... dh/dt = 32/(45π) m/s ≈ 0.22635 m/s
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You can also write an equation for the filled volume in terms of the filled height, then differentiate and solve for dh/dt. When you do, you find the relation between rates of change of height and area are as described above. We have taken a "shortcut" based on the knowledge gained from solving it this way. (No arithmetic operations are saved. We only avoid the process of taking the derivative.)
Note that the cone dimensions mean the radius is 3/8 of the height.
V = (1/3)πr²h = (1/3)π(3/8·h)²·h = 3π/64·h³
dV/dt = 9π/64·h²·dh/dt
.004 = 9π/64·0.2²·dh/dt . . . substitute the given values
dh/dt = .004·64/(.04·9·π) = 32/(45π)
The discriminant of this equation is 49. The solutions for this equation are x=2 and x=-
I hope that helps.
Step-by-step explanation:
5^5•5 = 5^3
y^2/y= y^1 =y
a^2•a^3•a= a^6
b^5/b^7= b^-2
y^5/y^4=y
m^3•m^5•m^2=m^10
Answer:
7x+ 14
Step-by-step explanation:
Answer:
-6
Step-by-step explanation:
By the segment addition postulate, RQ+QP=RP.
x+14+x+15=17
2x+29=17
2x = -12
x = -6
