Homework SolutionsMath1A Prof. OlssonFall 2008 Week 5Section 2.45.F (y) =1y2−3y4y + 5y3=−3y3+−14y+ 5yF0(y) =9y4+14y2+ 517.y =r21 +√rUsing the Quotient Rule,dydr=(1 +√r) 2r − r212√r(1 +√r)2=2r +32r√r(1 +√r)221.f (θ) =sec θ1 + sec θ=1cos θ + 1Using the Quotient Rule,f0(θ) =(cos θ + 1) · 0 − 1 · (−sin θ)(cos θ + 1)2=sin θ(cos θ + 1)225.f (x) =xx +cx=x2x2+ cUsing the Quotient Rule,f0(x) =(x2+ c) 2x − x2· 2x(x2+ c)2=2cx(x2+ c)229.The tangent line to the curve y = tan x at the pointπ4, 1has slopedydx= sec2π4= 1 + tan2π4= 21so using the point-slope equation for the line,y − 1 = 2x −π435.H (θ) = θ sin θUsing the Product Rule,H0(θ) = θ cos θ + sin θH00(θ) = −θ sin θ + 2 cos θ41.Using f (5) = 1, f0(5) = 6, g (5) = −3, and g0(5) = 2a)(fg)0(5) = f (5) g0(5) + f0(5) g (5) = 1 · 2 + 6 (−3) = −16b)fg0(5) =g (5) f0(5) − f (5) g0(5)g (5)2=−3 · 6 − 1 · 2(−3)2=−209c)gf0(5) =f (5) g0(5) − g (5) f0(5)f (5)2=1 · 2 − (−3) · 612= 2046.a)y = x2f (x) ⇒dydx= x2f0(x) + 2xf (x)b)y =f (x)x2⇒dydx=x2f0(x) − 2xf (x)(x2)2=xf0(x) − 2f (x)x3c)y =x2f (x)⇒dydx=2xf (x) − x2f0(x)f (x)2d)y =1 + xf (x)√x2dydx=√x (xf0(x) + f (x)) − (1 + xf (x))12√x√x2=2x2f0(x) + xf (x) − 12x√xSection 2.55.y =√u and u = sin x sodydx=12√ucos x =cos x2√sin x9.g (t) =1(t4+ 1)3g0(t) =−3(t4+ 1)4· 4t3=−12t3(t4+ 1)417.y = (2x − 5)48x2− 5−3dydx= (2x − 5)4h−38x2− 5−4· 16xi+ 4 (2x − 5)3· 2 ·8x2− 5−3= (2x − 5)38x2− 5−4−32x2+ 240x − 4035.y = cot2(sin θ)dydθ= 2 cot (sin θ) ·−csc2(sin θ) · cos θ39.The tangent line to the curve y = (1 + 2x)10at the point (0, 1) has slopedydx= 10 (1 + 2x)9· 2 = 20so using the point-slope equation for the line,y − 1 = 20 (x − 0)349.a)Using g (1) = 2, g0(1) = 6, and f0(2) = 5h (x) = f (g (x)) ⇒ h0(1) = f0(g (1)) · g0(1) = 5 · 6 = 30b)Using f (1) = 3, f0(1) = 4, and g0(3) = 9H (x) = g (f (x)) ⇒ H0(1) = g0(f (1)) · f0(1) = 9 · 4 = 3654.a)F (x) = f (xα) ⇒ F0(x) = f0(xα) · αxα−1b)G (x) = f (x)α⇒ G0(x) = α · f (x)α−1· f0(x)57.f (x) = 2 sin x + sin2xf0(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,which occurs at x =k +12π for integers k, or 1+sin x = 0, which occurs atx =k +12π for odd integers k. Therefore, the tangent lines passing throughthe pointsk +12π, 3for even integers k and the pointsk +12π, −1for odd integers k are horizontal.67.If θ is measured in degrees, then let ψ =π180θ be the corresponding radianangle. Using the Chain Rule,ddθsin ψ = cos ψ ·dψdθ=π180· cos ψSection 2.63.x3+ x2y + 4y2= 63x2+ 2xy + x2dydx+ 8ydydx= 04dydx=−3x2− 2xyx2+ 8y10.y sinx2= x siny2y cosx2· 2x + sinx2dydx= x cosy2· 2ydydx+ siny2dydx=2xy cos (x2) − sin (y2)2xy cos (y2) − sin (x2)11.tanxy= x + ysec2xy·y · 1 − xdydxy2= 1 +dydxdydx=yy2sec2xy− 1xy2sec2xy+ 1=y − y2cos2xyx + y2cos2xy19.The tangent line to the curve with equation x2+ y2= (2x2+ 2y2− x)2through the point0,12has slope given by2x + 2ydydx= 22x2+ 2y2− x4x + 4ydydx− 12 · 0 + 2 ·12dydx= 2 2 · 02+ 2122− 0!4 · 0 + 4 ·12dydx− 1dydx= 2 ·122dydx− 1dydx= 1Using the point-slope form of the equation of the line,y −12= 1 (x − 0)25.x3+ y3= 153x2+ 3y2dydx= 0 ⇒dydx=−x2y26x + 6ydydx2+ 3y2d2ydx2= 02x + 2y−x2y22+ y2d2ydx2= 0d2ydx2=−2x4y3− 2xy2=−2x4− 2xy3y532.The tangent line to the ellipsex2a2+y2b2= 1through the point (x0, y0) has slope given by2xa2+2yb2dydx= 0dydx=−b2x0a2y0Using the point-slope form for the equation of the liney − y0=−b2x0a2y0(x − x0)y0y − y20b2=−x0x + x20a2x0xa2+y0yb2=x20a2+y20b2= 135.For the curve with equation y = cx2, the slope is given bydydx= 2cxFor the curve with equation x2+ 2y2= k, the slope is given by2x + 4ydydx= 0 ⇒dydx=−x2y6The product of the two slopes at the point (x, y) is2cx−x2y=−cx2y=−yy= −1so the tangent lines to the two curves through the point (x, y) are perpen-dicular, and the two curves form orthogonal trajectories of each other.39.The ellipse with equation x2− xy + y2= 3 crosses the x-axis when y = 0, sox =√3 or x = −√3. The slope of the ellipse at point (x, y) is given by2x − y − xdydx+ 2ydydx= 0 ⇒dydx=2x − yx − 2yso the slope of the tangent line through√3, 0is 2. The slope of the tangentline through−√3, 0is also 2, so the lines are
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