What values of c and d make the equation true? Assume x greater-than 0 and y greater-than-or-equal-to 0 StartRoot StartFraction
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Given:
A bird was sitting 16 feet from the base of an oak tree and flew 20 feet to reach the top of the tree.
To find:
The length of the tree.
Solution:
A bird was sitting 16 feet from the base of an oak tree.
Vertical distance between bird and base = 16 feet
Then the bird flew 20 feet to reach the top of the tree.
Now, the vertical distance between bird and base = (16+20) feet
So, length of the oak tree is it the sum of 16 feet and 20 feet.
feet
feet
Therefore, the length of the oak tree is 36 feet.
At the start, the tank contains
(0.02 g/L) * (1000 L) = 20 g
of chlorine. Let <em>c</em> (<em>t</em> ) denote the amount of chlorine (in grams) in the tank at time <em>t </em>.
Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of
(<em>c</em> (<em>t</em> )/(1000 + (10 - 25)<em>t</em> ) g/L) * (25 L/s) = 5<em>c</em> (<em>t</em> ) /(200 - 3<em>t</em> ) g/s
In case it's unclear why this is the case:
The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after <em>t</em> s there will be (1000 + 10<em>t</em> ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time <em>t</em> is (1000 - 15<em>t </em>) L. Then the concentration of chlorine per unit volume is <em>c</em> (<em>t</em> ) divided by this volume.
So the amount of chlorine in the tank changes according to
which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5<em>t</em> )^(5/3), then integrate both sides to solve for <em>c</em> (<em>t</em> ):
There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :