Homework Solutions Math1A Prof Olsson Fall 2008 Week 5 Section 2 4 5 F y 1 3 4 2 y y 3 14 y 5y 3 3 5y y y F 0 y 17 y 9 14 2 5 4 y y r2 1 r Using the Quotient Rule dy dr 1 r 2r r2 2 1 r 21 f 1 2 r 2r 32 r r 2 1 r sec 1 1 sec cos 1 Using the Quotient Rule f 0 cos 1 0 1 sin sin 2 cos 1 cos 1 2 25 f x x x c x x2 x2 c Using the Quotient Rule f 0 x x2 c 2x x2 2x 2cx 2 2 2 x c x c 2 29 The tangent line to the curve y tan x at the point dy sec2 1 tan2 2 dx 4 4 1 1 4 has slope so using the point slope equation for the line y 1 2 x 4 35 H sin Using the Product Rule H 0 cos sin H 00 sin 2 cos 41 Using f 5 1 f 0 5 6 g 5 3 and g 0 5 2 a f g 0 5 f 5 g 0 5 f 0 5 g 5 1 2 6 3 16 b 0 g 5 f 0 5 f 5 g 0 5 3 6 1 2 20 f 5 2 2 g 9 g 5 3 c 0 1 2 3 6 g f 5 g 0 5 g 5 f 0 5 20 5 2 f 12 f 5 46 a y x2 f x dy x2 f 0 x 2xf x dx b y f x dy xf 0 x 2f x x2 f 0 x 2xf x x2 dx x3 x2 2 c y x2 dy 2xf x x2 f 0 x f x dx f x 2 d y 1 xf x x 2 dy dx x xf 0 x f x 1 xf x 2 1 x 2x2 f 0 x xf x 1 2 2x x x Section 2 5 5 y u and u sin x so dy 1 cos x cos x dx 2 u 2 sin x 9 g t g 0 t 1 t4 1 3 12t3 3 3 4t t4 1 4 t4 1 4 17 3 y 2x 5 4 8x2 5 h i 4 3 dy 2x 5 4 3 8x2 5 16x 4 2x 5 3 2 8x2 5 dx 4 2x 5 3 8x2 5 32x2 240x 40 35 y cot2 sin dy 2 cot sin csc2 sin cos d 39 The tangent line to the curve y 1 2x 10 at the point 0 1 has slope dy 10 1 2x 9 2 20 dx so using the point slope equation for the line y 1 20 x 0 3 49 a Using g 1 2 g 0 1 6 and f 0 2 5 h x f g x h0 1 f 0 g 1 g 0 1 5 6 30 b Using f 1 3 f 0 1 4 and g 0 3 9 H x g f x H 0 1 g 0 f 1 f 0 1 9 4 36 54 a F x f x F 0 x f 0 x x 1 b G x f x G0 x f x 1 f 0 x 57 f x 2 sin x sin2 x f 0 x 2 cos x 2 sin x cos x 2 cos x 1 sin x The tangent line will be horizontal when its slope is 0 so either cos x 0 1 which occurs at x k 2 for integers k or 1 sin x 0 which occurs at x k 12 for odd k Therefore the tangent lines passing through integers 1 the points k 2 3 for even integers k and the points k 21 1 for odd integers k are horizontal 67 If is measured in degrees then let angle Using the Chain Rule 180 be the corresponding radian d d sin cos cos d d 180 Section 2 6 3 x3 x2 y 4y 2 6 3x2 2xy x2 dy dy 8y 0 dx dx 4 dy 3x2 2xy dx x2 8y 10 y sin x2 x sin y 2 dy dy y cos x2 2x sin x2 x cos y 2 2y sin y 2 dx dx 2 2 dy 2xy cos x sin y dx 2xy cos y 2 sin x2 11 x tan x y y dy y 1 x dx x dy 2 sec 1 2 y y dx y 2 2 x 2 x y y cos sec 1 y2 y y dy x dx sec2 xy 1 x y 2 cos2 xy y2 19 2 The tangent line to the curve with equation x2 y 2 2x2 2y 2 x through the point 0 12 has slope given by dy dy 2 2 2x 2y 2 2x 2y x 4x 4y 1 dx dx 2 1 1 dy 1 dy 2 2 0 2 2 2 0 2 0 4 0 4 1 2 dx 2 2 dx dy 1 dy 2 2 1 dx 2 dx dy 1 dx Using the point slope form of the equation of the line y 1 1 x 0 2 25 x3 y 3 1 5 dy dy 0 dx dx 2 dy d2 y 6x 6y 3y 2 2 dx dx 2 2 2 x 2d y y 2x 2y y2 dx2 3x2 3y 2 x2 y2 0 0 4 2 xy3 2x d2 y 2x4 2xy 3 dx2 y2 y5 32 The tangent line to the ellipse x2 y 2 2 1 a2 b through the point x0 y0 has slope given by 2x 2y dy 2 0 a2 b dx dy b2 x0 2 dx a y0 Using the point slope form for the equation of the line y y0 b2 x0 x x0 a2 y 0 y0 y y02 x0 x x20 b2 a2 x0 x y0 y x20 y02 2 1 a2 b2 a2 b 35 For the curve with equation y cx2 the slope is given by dy 2cx dx For the curve with equation x2 2y 2 k the slope is given by 2x 4y dy dy x 0 dx dx 2y 6 The product of the two slopes at the point x y is x cx2 y 2cx 1 2y y y so the tangent lines to the two curves through the point x y are perpendicular and the two curves form orthogonal trajectories of each other 39 The ellipse with equation x2 xy y 2 3 crosses the x axis when y 0 so x 3 or x 3 The slope of the ellipse at point x y is given by dy dy 2x y dy 2y 0 dx dx dx x 2y so the slope of the tangent line through 3 0 is 2 The slope of the tangent line through 3 0 is also 2 so the lines are parallel 2x y x 7
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