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7 Pages

### m211feb25

Course: MATH 211, Fall 2009
School: Maryland
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Word Count: 1993

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2211 Math 3225 Lecture Feb 24-28 2003 Implicit Differentiation So far, all we've done is learn how to calculate derivatives of expressions of the form: y = f(x) ...in other words, when y was reduced to an explicit formula in terms of x. As you no doubt recall from algebra, it's quite rare that when presented with an equation involving y and x, we can hope to isolate y in terms of x. When we cannot do this,...

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2211 Math 3225 Lecture Feb 24-28 2003 Implicit Differentiation So far, all we've done is learn how to calculate derivatives of expressions of the form: y = f(x) ...in other words, when y was reduced to an explicit formula in terms of x. As you no doubt recall from algebra, it's quite rare that when presented with an equation involving y and x, we can hope to isolate y in terms of x. When we cannot do this, we're faced with a situation often described as a transcendental equation (an equation which transcends or goes beyond and algebraic method we can hope to use to solve it exactly.) Some examples (where in all cases, y depends on x): 1. 2. 3. sin(y + x) = x2y ycos(xy) + [tan(y 3)] = 10 csc[tan(xy) + y] = xy2 The question then arises, if we can't hope to obtain y explicitly in terms of x, can we think of obtaining an expression for y', or y's rate of change in terms of x? Yes! We simply differentiate across: 1.) d/dx sin(y + x) = d/dx (x2y) cos(y + x)y' = 2xy + x2y' cos(y + x)y' - x2y' = 2xy (Using the chain rule on LHS and Product Rule on RHS) [cos(y + x) - x2 ]y' = 2xy y'= 2xy . [cos(y + x) - x2 ] Though the equation is transcendental (we can't explicitly work out an exact expression for y in terms of x) we have obtained an exact formula for y's rate of change. One drawback is that this formula is in terms of y and x --and we don't know what y is in terms of x! (Hence the name "implicit derivative" ... we don't have the derivative as an explicit formula of x.) But this may not be such a drawback: oftentimes, even if we don't know y explicitly in terms of x, we have data at a point P(x,y), so we can plug in x and y in the above formula and know the rate of change of y at P. Think of it this way: given data (like in a scatter-plot) it's seldom we can easily construct explicitly the dependent variable Y in terms of the independent variable X. But it may not be necessary! We can get alot of information about the "trends" (y's rate of growth) 1 made possible by implicit differentiation. This is the true advantage and power of implicit differentiation. Equipped with implicit differentiation, we have converted our transcendental equation into a differential equation (an equation involving an unknown quantity y and its derivatives): cos(y + x)y' = 2xy + x2y' But there's more! Our explicit formula for y' also implies the necessary condition: cos(y + x) - x2 0 (the denominator can't equal zero.) This condition is called a constraint. Approaching a complex, dynamical problem by modelling it with a differential equation and a constraint is a very general and powerful application: it' practically the applied researcher's "Golden Rule." 2. d/dx ycos(xy) + [tan(y 3)] = d/dx 10 y'cos(xy) + y[-ysin(xy) - xy' sin(xy)] + 3y2y' sec2 (y3) = 0 Using Product Rule and chain rule for the cosine term Using Chain Rule y' [cos(xy) - xy sin(xy) + 3y2sec2 (y3)] = y2 sin(xy) y' = y2 sin(xy) . [cos(xy) - xy sin(xy) + 3y2sec2 (y3)] The differential equation Constraint: cos(xy) - xy sin(xy) + 3y2sec2 (y3) 0 3.) csc[tan(xy) + y] = xy2 d /dx csc[tan(xy) + y] = d/dx xy2 - csc[tan(xy) + y] cot[tan(xy) + y][sec2(xy)(xy' + x) + y'] = y2 + 2xyy' Make sure you understand all the steps in this procedure, (chain rule+product rule) y' {- csc(tan(xy) + y)cot(tan(xy) + y)sec2(xy)(x + 1) - 2xy} = y2 - csc(tan(xy) + y) cot(tan(xy) + y)sec2(xy) y' = - y2 + csc(tan(xy) + y) cot(tan(xy) + y)sec 2(xy) . csc[tan(xy) + y] cot[tan(xy) + y][sec2(xy)(xy' + x)] - 2xy Constraint: csc[tan(xy) + y] cot[tan(xy) + y][sec 2(xy)(xy' + x)] - 2xy 0 Observe that the constraint can be further simplified, it implies: csc[tan(xy) + y] 0 OR cot[tan(xy) + y] 0 OR [sec2(xy)(xy' + x) ] 0 OR 2xy 0 1 Which implies: tan(xy) + y n OR xy 0 (The OR is inclusive: meaning: "either/or and maybe both." 2 ) 1 2 Examples of inclusive OR. "Ketchup or mustard?" I can have either ketchup or mustard but also I can have both. On the other hand, "alive or dead?" is exclusive, can't be both, only either/or. 2 (Note that csc(..) and sec(..) is NEVER 0) This technique is known as implicit differentiation. Its adjective "implicit" also reveals its drawback: at best, we can crank out formulae for y'(x) that involve y(x) implicitly: y'(x) is written in terms of a formulae of y(x), and x, so we still don't know waht y is explicitly in terms of x. But sometimes we don't care (we don' t need to y(x) necessarily to find y'(x) at a point.) This is precisely implicit differentiation's advantage: we by-pass needing to know what y(x) is to find y'(x). Aside from this attractive feature, implicit differentiation may also be used to calculate y'(x) when the equation written in terms of y and x is solvable, but would involve a very cumbersome expression of y = f(x) and/or would take alot of time and energy to solve. For example, consider the follwing two problems: 1. Given the ellipse centered at (2,3) with semimajor axis length 6 (paralell to the xaxis) and seminor axis of length 4 (paralell) to y axis, derive y'. 2. Find y' given: x5 + 4xy3 - 3y5 = 5. Though both situations aren't transcendental, it would be tedious and cumbersome to first solve for y in terms of x explicily (especially in 2.). Use implicit differentiation instead: 1.) The equation of the ellipse is: Differentiating both sides: (x - 2) 18 y' = 9(x - 2 4(y ) - 3) (x - 2)2 + (y - 3)2 36 16 + (y - 3)y' = 0 8 = 1 This is easier to compute than having to first solve the equation for y and then find y': y' = d/dx { 3 + [16 - 4/9 (x - 2)2 ]1/2} = +1/2[16 - 4/9 (x - 2)2 ]-1/2[- 8/9 (x - 2) 2.) Find y' given: x5 + 4xy3 - 3y5 = 5. (The reader can complete this as an exercise) Implicit Differentiation can also be used to evaluate higher-order derivatives. Example: Calculate y" for: 1.) 2x3 - 3y2 =7 3 2.) x = sin2y, for P(1/2, /4) 1.) 2x3 - 3y2 =7 d /dx [2x3 - 3y2 ] = d/dx 7 3x2 - 6yy' = 0 y' = x2 2y d /dx [3x2 - 6yy'] = 0 6x - 6(y')2 - 6yy" = 0 y" = x - (y')2 y So we can This isn't the final answer, since we'd like y" expressed solely in terms of y and x. substitute the y' expression (necessarily derived first) y" = 4y2x - x4 4y2 2.) x = sin2y, for P(1/2, /4) d /dx x = d/dx sin2y 1 = 2sinycosy y' = y'sin2y y' = csc2y d /dx y' = d/dx csc2y y" = -2csc2ycot2y y' Since y' = csc2y then: y" = -2csc2 (2y)cot(2y) = -2cos(2y) . sin3 (2y) Extension of the Power Rule: Rational Exponents Recall that: xp/q = q(xp) = (qx)p (Proof: xp/q =xp*1/q =(xp )1/q= q(xp) = xp/q =x1/q *q =(x1/q )p= (qx)p ) Power-Rule (Rational Exponents) y = xp/q y' = p/q x[(p/q) - 1 ] Proof: yq = xp qyq-1y' = pxp-1 y' = p/qxp-1y1-q = p/qxpx-1yy-q= p/qxpx-1y(yq)-1 = p/qxpx-1y(xp)-1= p/qx-1y = p/qx-1xp/q= p/q x[(p/q) - 1 ] 4 APPLICATIONS (I) Word Problems Involving Instantaneous Rates of Change General Problem Solving Procedure: Example: At a certain instant the side of an equilateral triangle is Sketch Figure (If Necessary) centimeters long and is List all knowns increasing at k List all unknowns centimeters per minute. List all algebraic relations between knowns, unknowns How fast is area Differentiate Algebraic Relations (Implicitly, if need be) increasing? Let x be the length of each side. Then each angle is /3 and the height of triangle is: xsin(/3) = (3x)/2 x And area A = 1/2 xh = 3/4 x2 To find rate of increase of area (given x= and x'(t) = d dx /dt = k) differentiate A: /dt A(t) = d/dt 3/4 x2(t) = 3/4 d/dt x2(t) = dx 3 /4 2x(t) dx/dt = 3/2 x(t) dx/dt Substituting x= and x'(t) = : d/dt /dt = k : A (x=) = 3 /2 k Note on units: After deriving an expression it's a good habit to check consistency of units. Note that k is measured in terms of L/T (length/time) and in terms of L. Therefore the units of A' are in LL/T = L2/T . This is consistent since we're deriving an expression of the rate of change of area. Example: A snowball is shrinking in a manner such that its radius is shrinking at a contant rate, decreasing from 16 cm to 10 cm in 30 min. How fast is volume of snowball shrinking at instant when R=12cm? Similar to problem above, we need to first derive V'. Since the snowball is spherical, Then: d/dt V(t)= 4/3 r 3(t) V(t) = d/dt 4/3 r 3(t) = 4/3 d/dt r 3(t) = 4 /3 3r 2(t)r'(t) = 4 r 2(t)r'(t) Note that the above formula shows that V changes with respect to r according to: 4 r 2, according to the surface area! 5 To determine r'(t), we're told that the ballon shrinks at a constant rate, thus r varies linearly with respect to t. So we can simply compute the average rate of change of r = instantaneous...

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