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Chapter 3 101 - |:| PROBLEMS PLUS 1 Leta be the...

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Unformatted text preview: |:| PROBLEMS PLUS 1. Leta be the cv—coordinate of Q. Since the derivative of y 2 1 — 1:2 is y’ 2 —2m. the slope at Q is —2a. But since the triangle is equilateral. E/O—C 2 x/g/l. so the slope at Q is —\/5. Therefore. we must have that —2a 2 —\/§ 2 2> a 2 5?. Thus. the point Q has coordinates ($1 — (é) )_ (AC 1) and by symmetry P has coordinates (7%. i). 2.y2:1:323$+4 2> y’23w273.andy23(m2—x) 2> y: 3(2x7x) y' 2 6a: — 3. The slopes of the tangents of the two curves are equal ‘:'/ when 3x2 — 3 2 6a; — 3; that is, when :1: 2 0 or 2. At an 2 0. both _4" -5 tangents have slope —3. but the curves do not intersect. At x 2 2. "-gggfime v int rsect at 2,6 .80 both tangents have slope 9 and the cur es e ( ) y: X _ 3X+ 4 ~20 there is a common tangent line at (2, 6). y 2 93: ~ 12. 3. Let y 2 tan‘1 w. Then tan 3/ 2 as. so from the triangle we see that $ sin tarf1 ac 2 sin 2 ——. Usin this fact we have that 1 + Z i ) y m g x x . . h sin(tan_1(sinh 33)) 2 Ami — sm :3 2 tanhw.Hence \/ 1 + sinh2 ac — coshm 7 I sin—1(tanh cc) 2 sin‘1 (sin(tan_1(sinh a3))) 2 tan’1(sinh m). 4. We find the equation of the parabola by substituting the point (—100, 100). at which the car is situated. into the general equation 3/ 2 (1:132: 100 2 a(—100)2 2 a 2 fi. Now we find the equation ofa tangent to the parabola at the point (330.310). We can show that y' 2 a(2ar) 2 F10 (2.1:) 2 im. so an equation of the tangent is y — yo 2 $$0(:c — 2:0). Since the point (m0, yo) is on the parabola, we must have yo 2 fimg. so our equation of the tangent can be simplified to y 2 @363 + 5—105c0(m — (1:0). We want the statue to be located on the tangent line, so we substitute its coordinates (100 50) into this equation: 50— 2 100 —a:0 + —05x0(100 — $0) 2> 2% — 200160 + 5000 : 0 :> :50 : g [200 i W/2002 — 4 (5000)] :> 330 : 100 :t 50x5. But 200 < 100. so the car’s headlights illuminate the statue when it is located at the point (100 i 50 x/2. 150 — 100 x/2) % (29.3, 8.6). that is. about 29.3 111 east and 8.6 m north of the origin. 251 ...
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