(9/30/08)Math 10A. Lecture Example Sol utions for Instru ctorsSection 3.5. The trigonometric functions†Example 1 Find the derivativeddx(5 sin x − 6 cos x).Answer:ddx(5 sin x − 6 cos x) = 5 cos x + 6 sin xSOLUTION FOR INSTRUCTORS:ddx(5 sin x − 6 cos x) = 5ddx(sin x) − 6ddx(cos x ) = 5 cos x + 6 sin xExample 2 Find the derivative of y = sin(x1/2).Answer:dydx=ddx[sin(x1/2)] = cos(x1/2)ddx(x1/2) =12x−1/2cos(x1/2)SOLUTION FOR INSTRUCTORS:dydx=ddx[sin(x1/2)] = cos(x1/2)ddx(x1/2) =12x−1/2cos(x1/2)Example 3 What is the rate of change of y = x2sin x with respect to x at x = 5?Answer: [Rate of change] = 10 sin(5) + 25 cos(5)SOLUTION FOR INSTRUCTORS:y0=ddx(x2sin x) = x2ddx(sin x) + sin xddx(x2) = 2x sin x + x2cos x •y0(5) = 2(5) sin(5) + (52) cos(5) = 10 sin(5) + 25 cos(5)Example 4 Figure 1 shows the curve y = x cos x and its tangent line at x = π.Give an equation for the tangent line.xy2−2πFIGURE 1Answer: Tangent line: y = −xSOLUTION FOR INSTRUCTORS:y(x) = x cos x • y0= x(cos x)0+ cos x(x)0= −x sin x + cos x • y(π) = π cos(π) = −π •y0(π) = −1 • Tangent line: y = −π − (x − π) • y = −x†Lecture notes to accompany Section 3.5 of Calculus by Hughes-Hallett et al. All or most of the examples in these notes areexamples or ex ercises from Al Shenk’s calculus manuscript.1Math 10A. Lecture Example Solutions (9/30/08). Section 3.5, p. 2Example 5 Why are the tangent lines to the curve y = sin2x + 1 in Figure 2 allhorizontal?xy2y = sin2x + 1π 2πFIGURE 2Answer:ddx(sin2x + 1 ) = 2 sin x cos x is zero at x = 0, x =12π, x = π, x =32π and x = 2π.SOLUTION FOR INSTRUCTORS:ddx(sin2x + 1) =ddx[(sin x )2+ 1] = 2 sin xddx(sin x) = 2 sin x cos x is zero atx = 0, x =12π, x = π, x =32π and x = 2π.Example 6 Give an equation of the tangent line t o y = tan x at x =14π.Answer: Tangent line: y = 1 + 2(x −14π) • The curve and the tangent line are shown in Figure A7.xy114πy = tan xπFigure A7SOLUTION FOR INSTRUCTORS:f(x) = tan x • f0(x) =1cos2x• f (14π) = tan(14π) = 1 • f0(14π) =1cos2(14π)=11√22= 2• Tangent line: y = 1 + 2(x −14π). • The curve and the tangent line are shown in Figure A7.Section 3.5, p. 3 Math 10A. Lecture Example Solutions (9/30/08).Example 7 The length of the shadow cast on the horizontal ground by the 10-foot-highvertical pole in Figure 3 is s = 10 tan θ, where θ is the angle between thepole and the sun’s rays in the drawing. At what rate is the length of theshadow increasing when θ =13π if the angle θ is increasing112π radians perhour.10 f eetShadowSunθGroundsPoleFIGURE 3Answer:dsdt=103π feet per hou rSOLUTION FOR INSTRUCTORS:dsdt=ddt(10 tan θ) =10cos2θdθdt• Set θ =13π anddθdt=112π •dsdt=10cos2(13π)(112π) =10(12)2(112π) =4012π =103π feet per hourInteractive ExamplesWork the following Interactive Examples on Shenk ’s web page, http//www.math.ucsd.edu/˜ashenk/:‡Section 3.5: Examples 1 through 4Section 3.6: Example 1‡The chapter and section numbers on Shenk’s web s ite refer to his calculus manuscript and not to the chapters and sectionsof the textbook for the