Group+F+Period+6

**Section 4.6 Pages 251-252 Problems 6, 12, and 18** By Bryan Korona, Aidan Schenkus, and Andrew Edillo

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 * //**Area**//
 * If //a// and //b// are the le of two sides of a triangle and //θ// the measure of the included angle, the area //A// of the triangle is //A// = ( 1/2 ) //ab// sin //θ//. How is //dA/////dt// related to //da/////dt//, //db/////dt//, and //dθ/////dt//?
 * //**Solution**//
 * First, write out the equation given:
 * //A// = ( 1/2 ) //ab// sin //θ//
 * Take the first derivative of the equation given utilizing the chain rule and implicit differentiation.
 * //dA/////dt// = //d/////dt// (( 1/2 //ab// sin //θ// ))
 * Split the term within the parentheses into two parts to utilize the product rule.
 * //dA/////dt// = //d/////dt// (( 1/2 //ab// **·** sin //θ// ))
 * Take the first derivative. For simplicity's sake, let //dA/////dt//, //da/////dt//, //db/////dt//, and //dθ/////dt// to be //A//', //a//', //b//' and //θ',// respectively//.// For the term ( 1/2 //ab//), use the product rule, and because you are using implicit differentiation, use the chain rule.
 * //A//' = ( 1/2 //ab// **·** ( cos //θ// ) **·** //θ//' ) + (( 1/2 //a// **·** 1 **·** //b//' ) + ( //b// **·** 1/2 **·** 1 **·** //a//' ))( sin //θ// )
 * //A//' = ( 1/2 //abθ//' ( cos //θ// )) + (( 1/2 //ab//' ) + ( 1/2 //a//'//b// ))( sin //θ// )
 * You can either multiply out the "sin //θ//" or leave it as is. In either case, the answer is:
 * //dA/////dt// = ( 1/2 //ab// //dθ/////dt// ( cos //θ// )) + (( 1/2 a //db/////dt// ) + ( 1/2 //b da/////dt// ))( sin //θ// )
 * //dA/////dt// = ( 1/2 //ab// //dθ/////dt// ( cos //θ// )) + ( 1/2 //a db/////dt// )( sin //θ// ) + ( 1/2 //b da/////dt// )( sin //θ// )
 * //**Growing Raindrop**//
 * Suppose that a droplet of mist is a perfect sphere and that, through condensation, the droplet picks up moisture at a rate proportional to its surface area. Show that under these circumstances the droplet's radius increases at a constant rate.
 * **//Solution//**
 * Draw a diagram.
 * List your rates. You want to find and prove that the rate "//r//" increases at a constant rate given that the rate at which the droplet picks up moisture (or, increases in volume) is proportional to the rate at which surface area increases.
 * dr/dx = ?
 * dv/dx = ?
 * da/dx = ?
 * List your equations:
 * Surface Area = A = 4 πr 2
 * Volume = V = ( 4/3 ) πr 3
 * y = kx; allowing "y" to equal "dv/dx" and x "da/dx," k being the rate at which "r" increases
 * Derive:
 * Surface Area:
 * da/dx = d/dx(4 πr 2 )
 * da/dx = 8πr(dr/dx)
 * Volume:
 * dv/dx = d/dx ( 4/3 ) πr 3
 * dv/dx = 4 πr 2 (dr/dx)
 * Isolate dr/dx:
 * (da/dx)/ 8πr = dr/dx
 * (dv/dx)/4 πr 2 = dr/dx
 * Set both rates equal to each other:
 * (da/dx)/ 8πr = (dv/dx)/4 πr 2
 * Cross multiply:
 * (da/dx)4 πr 2 = (dv/dx) 8πr
 * Divide to discover da/dv, in other words, the proportion of the rate of increase between area and volume:
 * (da/dx)4 πr 2 /(dv/dx) 8πr = 1
 * (da/dv)1r/2 = 1
 * 1r/2 = dv/da
 * Notice the "r." Your solution is 1/2, but what it's saying is that as "r" increases, surface area increases twice as fast as volume, as the solution can be rewritten as "1r : 2"
 * //**Draining Hemispherical Reservoir**// Water is flowing at the rate of 6m3/min from a reservoir shaped like a hemispherical bowl of radius 13 m, shown here in profile. Answer the following questions given that the volume of water in a hemispherical bowl of radius R is V = (π/3)y2(3R-y) when the water is y units deep.
 * //**Part A**// At what rate is the water level changing when the water is 8 m deep?
 * //**Solution**//:
 * Utilize diagram given:
 * Find out what rates you must use. You must solve for dy/dt, as y represents the water level and //dy/////dt// represent the rate at which it decreases.
 * //dy/////dt// = ?
 * Write out the equation given, then substitute in 13 for R, as the as capital R/radius is given to you.
 * //V// = ( π/3)//y// 2 (3(13(R)-y)
 * //V// = ( π /3)//y// 2 (39-y)
 * Take the first derivative of the equation and then substitute the other information given to you. Solve for dy/dt.
 * //y// = 8 m
 * //dV/////dt// = -6 m 3 /min
 * //dV/////dt// = ( π /3 )( y 2 ( -//dy/////dt// ) + ( 39-//y// )( 2//y// ( //dy/////dt// ))
 * -6 = ( π /3 ) ( 82 ( -//dy/////dt// ) + ( 39-8 )( 2//y// ( //dy/////dt// ))
 * -6 = ( π /3) ( -64 ( //dy/////dt// ) + 496 ( //dy/////dt// ))
 * -18/ π = -64 ( //dy/////dt// ) + 496( //dy/////dt// )
 * -18/ π = 432 ( //dy/////dt// )
 * dy/dt = ( -18/ π ) / 432
 * dy/dt ≈ -0.01326 m/min
 * **//Part B//** What is the radius r of the water's surface when the water is y m deep?
 * //**Solution**//
 * Solve for r, realizing the shape your are dealing with is a triangle, with sides "13 - y", "r", and a hypotenuse of 13. Use the Pythagorean theorem.
 * Side 2 = 13 - //y//
 * ( Side 2 ) 2 + //r// 2 = 13 2
 * ( 13 - //y// ) 2 + //r// 2 = 169
 * 169 - 26//y// + //y// 2 + //r// 2 = 169
 * -26//y// + //y// 2 + //r// 2 = 0
 * //r// 2 = 26y - //y// 2
 * //r// = ( 26//y// - //y// 2 ) 1/2
 * //**Part C**// At what rate is the radius r changing when the water is 8 m deep?
 * //**Solution**//
 * You found r as it relates to y in part B. Find the first derivative of the solution to part B.
 * //r// = ( 26y - //y// 2 ) 1/2
 * //dr/////dt// = ( 1/2 ) ( 26y-//y// 2 ) -1/2 ( 26-2//y// ) ( //dy/////dt// )
 * Substitute the answer from part A into //dy/////dt//
 * //dr/////dt// = ( 1/2 ) ( 26(8)-82 ) -1/2 (26 - 2 ( 8 )) (( -18/ π ) / 432 )
 * //dr/////dt// = -0.0055262 m/min
 * ** π **