The derivative of the function p(t) = (a – bt)2, where a and b are constants, is p' (t) = 2b2t – 2ab. For example, if a
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The derivative of the function p(t) = (a – bt)2, where a and b are constants, is p' (t) = 2b2t – 2ab. For example, if a
The derivative of the function p(t) = (a – bt)2, where a and b are constants, is p' (t) = 2b2t – 2ab. For example, if a = 2 and b = 5, then the derivative of p(t) = (2 – 5t)2 is p'(t) = 50t – 20. = Large cylindrical tanks are used in industrial settings for storing large quantities of chemicals in liquid form. For various reasons, e.g. maintenance, tanks need to be drained on a regular basis. This is done by opening a draining valve at the bottom of the tank and letting the liquid slowly escape through the draining pipe. Using the equations for modelling fluid flow, we can show that during the draining process, the height of liquid in the tank depends on time according to the formula: p(t) = (vho – kt)?, = where ho is the height of liquid in the tank at the start of the draining process when t = 0. Here t is expressed in SI units. The constant k is given by the formula a k = A g 2 where a is the area of the draining hole, A is the cross-sectional area of the tank and g= 9.81 m s-2 is the acceleration due to gravity. A ho a
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