Question

IUIUI LULUULIV KELLSHM 15 June - 21 June / MATH 1700 - Summer 2 Let the parametric curve C be defined by x = 1 - ln(t + 4), y = 1 + In(4-1), for all – 4 < t < 4. a) [4.5] Find the equation of the line tangent to Cat = 1. b) [4.5] Does the curve C have a horizontal tangent line? If yes, find the coordinates of the point(s) at which C has horizontal tangent(s). If no, explain why. C) (4.5] Does the curve C has a vertical tangent line? If yes, find the coordinates of the point(s) at which C have vertical tangent(s). If no, explain why. day d) (4) Find Do not simplify your answer. dx² 1 A -

Answer 1

Soly na it- luft+4) 'ga ttln (Art) - 44t24 1X(0+1) 1 4 tt 4ttn 4 ++) (4+3) (4+4) A dd 7 t 1 & -t x (only dut = 4-tal - (+3) 9-t dy dyldt 1 dn (art) (4+4) (4-4) (3+1) dalat tai 2 (5) = da (3-1) (441) (4-1) (3+0) 3(4) 6 dy da ho and slope of the tangent in (m) = X(1) Tolm(5) H)= If ln(3) How, equation of the tangent line is y-y (1) man (1) 1-1*13) = 5 Y-12 6 (4- Itln(5)) Ex-a + Elml5) +1+8113)

{ x + In13) + ln 13) + = ln (5) + 7 Horizontal fangents) (ort) 3-tao ㅠ (4-34+0) tr3 x(3) = 3-2017) , y 13)= 3+ ln (D - 3toa 3 QD) = (3-2017), 3) Vertical tangents ++3 & t3 ao t +4 (t+4+0) + -3 3 *(-3)= -3- fu(1) = -300 8 -3) = -3 + ( 4 + 3) 3+ ln (7) (x,y)= (-3, -34 ln(7)

412 da (3) : (4) (一) ( # 6-4+4) 4 - 4/3++) ( 4X (43) 1444) d} 一 C22-24) x 4+4) (4-〉++) (3++) da 3 div> -3.57-8-144 (4 -1)*(2+1)

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