<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
_____
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)
Im not sure, but i think it should be
( -3, 1 )
( -3, -2 )
( -6, -3 )
A = $12105; P = $19400; t = 3 (yrs)
Then $12105 = $19400(1-r)^3, or
0.6240 = (1-r)^3
Take the cube root of both sides. Then 0.854 = 1 - r, or r = 1 - 0.854 = 0.145 (answer) 0.145 is a decimal fraction.
Answer:
The equation of the tangent at x=-6 is
Step-by-step explanation:
The equation of a circle with center (h,k) with radius r units is given by:
The given circle has center (-3,1) and radius 5 units.
We substitute the center and the radius into the equation to get;
To find the slope, we differentiate implicitly to get:
When x=-6;we have
or
From the graph the reuired point is (-6,-3).
We substitute this point to find the slope;
The equation is given by .
We plug in the slope and the point to get: