Question 1

Recall the formula for tan(a + b) in terms of tan(a) and tan(b): tan(a,) +tan(b) \"mm + b) = 1tan(a,) tan(b) ' Now use this to write tan1(m) + tan1 (y) = tan1(0) where 6 = E. What might be an interpretation of this if we try to substitute y = % for :z: > 0? Since this leads to a denominator of Number we conclude that the sum tan1(3) + tan_1(%) = E! E. Curves are usually specified in terms of a single equation. For example the equation of a circle of radius 1 is given by 2 2 ty2 = 12 Every point [x, y] on the curve satisfies this relation. Alternatively, points on the curve can be specified in terms of a parameter u: [cos u, sinu] 0 Sus 27 . Another possibility is the rational parametrization: a (t ) = 1-+2 2t tER 1+ + 2 ' 1 + + 2 which allows you to find points on the unit circle with infinite precision by substituting t values. For example . a(0) = (1,0) . a(-3) = (-4/5,-3/5) . a (-2) = (4/5,-3/5) For a general parametristd curve [z (t) , y (t)], the slope of the tangent at some point when t = c is given by dy (c ) at (c ) . dx dy (c) = dt And the slope of the normal line at some point when t = c is the negative reciprocal of the slope of the tangent line, namely dx (c) = _dx (c) at (c) . dy Hence, using the rational parametrication of the unit circle above dy (-3) = Number dt dac (-3) = Number dt The slope of the tangent line to the curve at [-4/5, -3/5] is- dy ( -3) = Number dx . The slope of the normal line to the curve at [-4/5, -3/5] is - dac (-3) = Number dyGiven the rational parametrication of the unit circle from the previous question 1-+ 2 a(t) : 2t 1+t2 ' 1+t2. Move the t slider in the GeoGebra app below to t = -3 . 1.5 t = -3.4 0.5 -2 -1.5 -0.6 0.5 1.5 2 -0.5 - a(-3.4) 1.5 -2 The equation of the tangent (displayed above in blue) at a(-3) is y = The equation of the normal (displayed above in red) at a(-3) is y =Consider the curve with parametric equation a(t) = [3t+3,3t2 +2t+ 1] ,t e R. The equation of the: . tangent to the curve at the point (1(1) is y = Q E] . normal to the curve at the point (1(1) is y : [Q B}. By eliminating the parameter t , we nd that the Cartesian equation of the curve is: y = E]- The curve (in black), tangent (in blue) and normal (in red) are displayed below. Let's return to the Folium of Descartes x3 + y3 = 3mg . This curve has rational parametrisation 3t 3t2 am = [_ _l . 1+1:3 ' 1+1:3 Set if = 4 in the GeoGebra app below. --3xy+y3=0 3 t .43 l' , _. " ' (-4-8) 2 _3 Find the point (1(4) = [12!65,48/65] @ :1. The equation of the tangent to the curve at the point (1(4) is y 2 IQ 1. The equation of the normal to the curve at the point (1(4) is y = @ :1. Note: the Maple syntax for the point [1, 2] is [1 , 2]