MTH405_Wk_8_Lect_1

# MTH405_Wk_8_Lect_1 - MTH405 Wk 8 Lect 1 Product Rule The...

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Unformatted text preview: MTH405 Wk 8 Lect 1 Product Rule The Product Rule deals with functions multiplied together, i.e., y = f (x) Â· g(x). Theorem. For multiplication of two functions have y = f (x) Â· g(x), we d [f (x) Â· g(x)] = [f 0 (x) Â· g(x)] + [g 0 (x) Â· f (x)] dx or in a more usable form y 0 = f 0 g + g 0 f. Quotient Rule The Quotient Rule deals with functions divided with each other, i.e., y= f (x) g(x) where g(x) 6= 0. Theorem. For division of two functions f (x) , we have g(x)   d f (x) f 0 (x)g(x) âˆ’ g 0 (x)f (x) = dx g(x) (g(x))2 or in a more usable y0 = f 0g âˆ’ g0f . g2 1 Chain Rule The Chain Rule deals with functions composed inside functions, i.e., y = f (g(x)). Theorem. For composition of two functions y = f (g(x)), we denote u = g(x), then we have y = f (u), hence the Chain Rule is dy dy du = Â· . dx du dx Example. Dierentiate these functions. 1. y = x sin x = (x)(sin x) 2. f (x) = x2 ex = (x2 ) (ex ) 3. f (x) = (x2 + 3x)(x3 + 30x2 + 7x âˆ’ 4). 4. y= x sin x ex 5. y= 2x+3 5x+10 6. y= ex x2 7. y = sin x2 (u = x2 is the inside function and y = sin u is outside , function), 8. y = (x2 âˆ’ 40)20 (u = x2 âˆ’ 40 is the inside function and y = u20 9. y = e5x (u = 5x is the inside function and y = eu is the outside is outside function ), function), 2 Implicit Dierentiation Till now, we have been dierentiating functions which look nice and organised, i.e., we have y on the left equals to a function on the right expressed in terms of x. So we notice the function and choose a rule to use. But what about this function? y Â· tan3 (xy 2 + y) = x Ofcourse we can not use any of the rules since y is also a part of the function. So our question is: How do we dierentiate functions which are very complicated and x and y variables are jumbled up? We use the technique of implicit dierentiation. â€¢ â€¢ Explicit dierentiation is where in the function, you can separate the variables x and y . And y is alone on the left of the equal sign and the rest of the function in terms of x is on the right of the equal sign. Till now, we have covered all the aspects of explicit dierentiation. x2 +x Example, y = sin x, y = 3x 2 +x and so on. In explicit dierentiation, we only dierentiate the function in terms of x. Implicit dierentiation is where in the function, you can not separate the variables x and y (or by separating them, the function becomes too complicated to use). Then both x and y are to be dierentiated. Note that whenever any y term is dy dierentiated, then multiply it with dx . Example. Use the technique of implicit dierentiation to nd 1. 5y 2 + sin y = x2 . 2. xy = 1. 3. x3 + y 3 = 3xy . 4. xy 3 1+sec y = 1 + y4. 3 dy dx . Applications of Dierentiation Rates of Change - Distance, Velocity and Acceleration Given a function which gives the distance travelled by an object moving in a direction is denoted by y = s(t) = d(t) = x(t). Note that the distance function is a function depending on t time. Then the velocity function is given by v(t) = and the acceleration dy = s0 (t) = d0 (t) = x0 (t) dt function is given by   d2 y d dy = 2 = s00 (t) = d00 (t) = x00 (t). a(t) = v (t) = dt dt dt 0 The velocity function v(t) = x0 (t) gives the instantanous velocity of the object at time t and the acceleration function a(t) = v 0 (t) gives the instantaneous acceleration of the object at time t. If y = x(t), then we dene the average velocity (average rate of change of y with respect to t) over the interval [t0 , t1 ] to be v ave = x(t1 ) âˆ’ x(t0 ) . t1 âˆ’ t0 4 Supplies are dropped from a helicopter and the distance fallen in a time t seconds is given by: Example. 1 x(t) = gt2 , 2 where g = 9.8m/s2 1. Determine the velocity function of the supplies. 2. Determine the acceleration function of the supplies. 3. 4. 5. 6. Determine the velocity and acceleration of the supplies after it has fallen for 2 seconds? After how many seconds does it hit the ground if the supplies are dropped from 100m? What is the velocity of the supplies as it hits the ground? What is the average velocity of the supplies? âˆ’xi over the interval [ti , tf ]). (Hint: vave = xtff âˆ’t i Solution. 1. Direntiating x(t) once, we have 1 2 gt 2 v(t) = x0 (t) = gt x(t) = 2. Dierentiating v(t) once, we have v(t) = gt a(t) = g = 9.8m/s2 . 3. Finding v(2) and a(2), we have v(2) = g(2) = (9.8)(2) = 19.6m/s a(2) = g = 9.8m/s2 . 5 4. Using the distance function to nd time of free fall, we have 1 2 gt 2 1 100 = (9.8)t2 2 200 2 t = 9.8 t = 4.516 sec. x(t) = 5. Since it takes t = 4.516sec to hit the ground, we input it in the velocity function v(t) = gt v(4.516) = (9.8)(4.516) = 44.257m/s. 6. Average velocity x(tf ) âˆ’ x(ti ) tf âˆ’ ti 100 âˆ’ 0 = (4.516) âˆ’ 0 = 22.14m/s. vave = 6 Optimization (Maximization and Minimization) Note that Optimization involves nding turning points to maximize or to minimize. A rectangular area is formed having a perimeter of 40cm. Determine the length and width of the rectangle such that it encloses the maximum area. Example. Solution. This is a maximization question since we are maximizing area. So let see what we know about this rectangle (Note that in this question we can use all 4 sides of the rectangle). Perimeter P = 40 = 2L + 2W Area A = L Â· W Our aim is to nd the values of L and W such that area is maximised. So we use perimeter equation, and make W the subject. So 40 = 2L + 2W W = 20 âˆ’ L and substitute into area equation A = LÂ·W = L(20 âˆ’ L) A = 20L âˆ’ L2 Then dierentiating and then equating to 0, we have dA = 20 âˆ’ 2L dL 0 = 20 âˆ’ 2L L = 10cm. Then W = 10cm. 7 Determine the area of the largest piece of rectangular ground that can be enclosed by 100m of fencing, if part of an existing straight wall is used as one side. Example. Solution. Note that in this question we can use 3 sides of the rectangle since 1 side is backed by the wall so it does not require fencing. So let see what we know about this rectangle. Perimeter P = 100 = L + 2W Area A = L Â· W Our aim is to nd the values of L and W such that area is maximised. So we use perimeter equation, and make W the subject. So 100 = L + 2W 1 W = 50 âˆ’ L 2 and substitute into area equation A = LÂ·W 1 = L(50 âˆ’ L) 2 1 2 A = 50L âˆ’ L 2 Then dierentiating and then equating to 0, we have dA = 50 âˆ’ L dL 0 = 50 âˆ’ L L = 50cm. Then W = 50 âˆ’ 21 L = 50 âˆ’ 21 (50) = 25m. 8 ...
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