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Unit #5 - Implicit Differentiation, Related Rates

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Unit #5 - Implicit Differentiation, Related Rates Unit #5 - Implicit Differentiation, Related Rates Some problems and solutions selected or adapted from Hughes-Hallett Calculus. Implicit Differentiation 1. Consider the graph implied by the equation (c) Setting the y value in both equations equal to xy2 = 1. each other and solving to find the intersection point, the only solution is x = 0 and y = 0, so What is the equation of the line through ( 1 ; 2) 4 the two normal lines will intersect at the origin, which is also tangent to the graph? (x; y) = (0; 0). Differentiating both sides with respect to x, Solving process: dy 3 3 y2 + x 2y = 0 (x − 4) + 3 = − (x − 4) − 3 dx 4 4 3 3 so x − 3 + 3 = − x + 3 − 3 4 4 dy −y2 3 3 = x = − x dx 2xy 4 4 6 x = 0 4 dy At the point ( 1 ; 2), = −4, so we can use the x = 0 4 dx point/slope formula to obtain the tangent line y = −4(x − 1=4) + 2 Subbing back into either equation to find y, e.g. y = 3 (0 − 4) + 3 = −3 + 3 = 0. 2. Consider the circle defined by x2 + y2 = 25 4 3. Calculate the derivative of y with respect to x, (a) Find the equations of the tangent lines to given that the circle where x = 4. (b) Find the equations of the normal lines to x4y + 4xy4 = x + y this circle at the same points. (The normal line is perpendicular to the tangent line.) Let x4y + 4xy4 = x + y. Then (c) At what point do the two normal lines in- dy dy dy 4x3y + x4 + (4x) 4y3 + 4y4 = 1 + tersect? dx dx dx hence Differentiating both sides, dy 4yx3 + 4y4 − 1 = : dx 1 − x4 − 16xy3 d d x2 + y2 = (25) dx dx 4. Calculate the derivative of y with respect to x, dy given that 2x + 2y = 0 dx xey = 4xy + 5y4 dy −2x −x = = dx 2y y dy To solve for dx , we must think of y as a function of x (a) The points at x = 4, satisfying x2+y2 = 25, would and differentiate both sides of the equation, using the be y = 3 and y = −3. chain rule where appropriate: Using the point/slope formula for a line, and our dy dy dy dy ey + xey = 4y + 4x + 20y3 calculated , through (4, 3): y = −1:33333 ∗ dx dx dx dx (x − 4) + 3, and dy Now, we simplify and move the terms with a dx to the through (4, -3): y = 1:33333 ∗ (x − 4) + (−3) dy right, and keep the terms without a dx to the left: (b) If you have a line with slope m, the slope of a ey − 4y = (4x + 20y3 − xey) dy perpendicular line will be −1=m. dx Finally, we can solve for dy : 3 dx y Through (4, 3): y = 4 (x − 4) + 3, and dy e − 4y through (4, -3): y = 3 (x − 4) + (−3) = 4 dx 4x + 20y3 − xey 1 To simplify a little we multiply both sides through by 5. Use implicit differentiation to find the equation p of the tangent line to the curve xy3 + xy = 14 2 x + y: at the point (7; 1). dy p p 1 − 4x2y x + y = 4xy2 x + y − 1 dx d To get the slope, we take the derivative of both p dx dy (4xy2 x + y) − 1 = p sides. dx 1 − 4x2y x + y d d xy3 + xy = (14) 7. Find all the x-coordinates of the points on the dx dx curve x2y2 + xy = 2 where the slope of the tan- dy dy (1)y3 + x 3y2 + (1)y + x = 0 gent line is −1. dx dx dy We need to find the derivative by implicit differen- dy dx Gathering terms with , and those without it, tiation. Differentiating with respect to x on both sides dx of the equation, dy 3xy2 + x + y3 + y = 0 dy dy dx 2xy2 + x2 2y + y + x = 0 dx dx dy y3 + y = − dx 3xy2 + x dy Here, we could solve for , but we actually know the dx dy slope we want this time: = −1, so let's just sub At the point (x; y) = (7; 1), dx that in now: dy 13 + 1 = − 2xy2 + x2 (2y(−1)) + y + x(−1) = 0 dx 3(7)(1)2 + 7 2 2 −2 −1 or 2xy − 2x y + y − x = 0 = = 28 14 factoring first terms: 2xy(y − x) + (y − x) = 0 Factoring common y − x:(y − x)(2xy + 1) = 0 To build a line that goes through the point (7, 1), and −1 with slope , we can use the point-slope line formula. Meaning either y − x = 0 (so y = x), or (2xy + 1) = 0. 14 We obtain the formula for the tangent line: Subbing each possibility into the equation for the curve, x2y2 + xy = 2 −1 y = (x − 7) + 1 14 If y = x, x2(x2) + x(x) = 2 x4 + x2 = 2 6. Find dy=dx by implicit differentiation. x4 + x2 − 2 = 0 p 2 2 x + y = 9 + x y (x2 + 2)(x2 − 1) = 0 d So x = −1; 1 are two solutions, with the corresponding Taking of both sides, dx y values, using y = x being y = −1 and 1 as well. d p d The other possible case, 2xy + 1 = 0, leads to x + y = 9 + x2y2 dx dx 2 −1 −1 1 1 dy dy If y = −1=2x, x2 + x = 2 p 1 + = 2xy2 + x22y 2x 2x 2 x + y dx dx 1 −1 or + = 2 Expanding left side: 4 2 1 1 dy 1 1 dy p + p = 2xy2 + x22y 2 x + y dx 2 x + y dx which is impossible, so the assumption that y = −1=2x must be impossible to use with this curve. dy Gathering terms with and without , 2 2 dx Therefore the only two points on the curve x y +xy = dy 2 which have slope = −1 are (x; y) = (1; 1) and dy 1 1 dx p − 2x2y = 2xy2 − p dx 2 x + y 2 x + y (−1; −1). 2 8. Where does the normal line to the ellipse x2 − xy + y2 = 3 at the point (−1; 1) intersect the ellipse for the second time? To obtain a normal (perpendicular) line, we find a line perpendicular to the tangent line on the ellipse at the point (-1, 1). (Linear algebra students may have other ways to do this.) To get the slope of the tangent line, we use implicit differentiation, with respect to x: d d 9. The curve with equation 2y3 + y2 − y5 = x4 − x2 − xy + y2 = (3) dx dx 2x3 + x2 has been likened to a bouncing wagon dy dy (graph it to see why). Find the x-coordinates 2x − y + x + 2y = 0 dx dx of the points on this curve that have horizontal tangents. dy (−x + 2y) = −2x + y dx dy −2x + y d = Taking an implicit of both sides, dx (−x + 2y) dx d d 2y3 + y2 − y5 = x4 − 2x3 + x2 dx dx At the point on the ellipse (−1; 1), dy dy dy 6y2 + 2y − 5y4 = 4x3 − 6x2 + 2x dy −2(−1) + 1 dx dx dx = dx (−(−1) + 2(1)) = 3=3 = 1 dy Since we know we want points where dx = 0 (horizon- dy So the slope of the tangent line to the ellipse at (−1; 1) tal tangents), we'll set dx = 0 immediately: is 1. 6y2(0) + 2y(0) − 5y4(0) = 4x3 − 6x2 + 2x The slope of the normal line will be perpendicular to 2 that, or −1=(1) = −1. factoring: 0 = 2x(2x − 3x + 1) 0 = 2x(2x − 1)(x − 1) The normal line, which also passes through (-1, 1), will therefore be 1 y = −1(x − (−1)) + 1 So x = 0; 2 and 1 are the points where the tangent to or y = −1(x + 1) + 1 the graph will be horizontal. or y = −x Note that the question only asked for the x coordinate of these points. As you can see on the graph below, all of these x coordinates correspond to multiple y values, To find the intersections of this normal line with the el- but all of those points have horizontal tangents. lipse to find a second crossing, we sub in this expression for y into the ellipse formula: x2 − xy + y2 = 3 x2 − x(−x) + (−x)2 = 3 x2 + x2 + x2 = 3 x2 = 1x = ±1 Since x = −1 is the point we started at, x = +1 must be the other intersection. At that point, y = −x, so the point is (x; y) = (1; −1). Below is a graph of the scenario. 3 10. Use implicit differentiation to find an equation Substituting x = 1 back into the circle equation to of the tangent line to the curve find the matching y coordinates gives y2(y2 − 4) = x2(x2 − 5) (1)2 + y2 − 2(1) − 4y = −1 y2 − 4y = 0 at the point (x; y) = (0; −2). y(y − 4) = 0 y = 0; 4 The points with horizontal tangents are (1,0) and (1,4). (b) The points with vertical tangents are those where dy the denominator of dx is zero (making the slope undefined). From part (a), we have dy −2x + 2 = dx 2y − 4 (The devil's curve) Setting the denominator equal to zero gives This point happens to be at the bottom of the loop on 2y − 4 = 0 the y axis, so the slope there is zero. Therefore, the tangent line equation is simply: y = 2 y = −2 Substituting y = 2 back into the circle equation to find the matching x coordinates gives 11.
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