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Section 2 Parametric Differentiation Theorem Let x = f(t), y = g(t) and dx/dt is nonzero, then dy/dx = (dy/dt) / (dx/dt) ; provided the given derivatives exist Example 1 Let x = 4sint, y = 3cost. Find: 1. dy/dx and d2y/dx2 2. dy/dx and d2y/dx2 at t = π/4 3. Find the slope & the equation of the tangent at t = π/4 Solution- Part1 dx/dt = 4cost, dy/dt = -3sint → dy/dx = (dy/dt) / (dx/dt) = - 3sint / 4cost = -(3/4)tant d(dy/dx)/dt = -(3/4)sec2t d2y/dx2 = d(dy/dx )/dx= [d(dy/dx)/dt] / [dx/dt] = [-(3/4)sec2t ] / 4cost = (-3/16) sec3t (dy/dx)( π/4) = -(3/4)tan( π/4) = -(3/4) (d2y/dx2 )( π/4) = (-3/16) sec3( π/4) = -(3/16)(√2) 3 = - 3√2/8 Part 2 The slope of the tangent at t = π/4 is equal to the derivative dy/dy at t = π/4, which is -(3/4). The Cartesian coordinates of the point t = π/4 are: x = 4sint π/4 = 4(1/√2)= 2 √2 and y = 3cos π/4= 3(1/√2)= (3/2) √2 The equation of the tangent at t = π/4 is: y -(3/2) √2 = -(3/4) ( x - 2 √2 ) Example 2 Let x = 4sint, y = 3cost. Find: 1. dy/dx and d2y/dx2 at the point (0, -3 ) 2. The equation of the tangent to the curve at that point. Solution – Part 1 . Let x = 4sint, y = 3cost. First we find any value of t corresponding to the point (0, -3 ). It is clear that one such value is t = π. Why?* Now, we substitute that in the formulas of dy/dx and d2y/dx2 , which we have already deduced in the Example(1) (dy/dx)( π) = -(3/4)tan( π) = 0 (d2y/dx2 )( π) = (-3/16) sec3( π) = -(3/16)(-1) 3 = 3/16 *Answering “Why?” We have: 0 = 4sint, -3 = 3cost. → sint = 0 & cost = -1 → t = ………,-3π, -π, π, 3π, 5π,….. Notice that the values for any trigonometric function at any of these numbers (angles) are the same. Take: t = π Solution – Part 2 From the slope of the tangent at (0 ,-3 ) it is clear that the tangent is horizontal, and hence it’s equation is: y = -3 We could also get that from the straight line’s formula: y – (-3)= ( x -0 ) → y + 3 =0 →y=- 3 But this is not very smart. It is like catching a fly with a hammer! Book Example (1) Let x =t2, y = t3 - 3t. 1. Find the equations of all tangents at (3,0) 2. Determine at which point (points), the graph has a horizontal tangent. 3. . Determine at which point (points), the graph has a vertical tangent. * 4. . Determine when the curve is concave upward / concave downward Solution-Part 1 We have: dx/dt = 2t , dy/dt = 3t2 -3 dy/dx = (dy/dt) / (dx/dt) = (3t2 -3) / t When x =3 & y=0 → 3=t2→ t=√3 or t= -√3 At t=√3, we have dy/dx=(9-3)/(2√3 )= √3 At t= -√3, we have dy/dx=(9-3)/(-2√3 )= -√3 Thus, the equations of the tangents to the curve at (x,y) = (3,0) are: y - 0 = √3 (x - 3) & y - 0 = -√3 (x - 3) That’s: y = √3 (x - 3) & y = -√3 (x - 3) Solution-Part 2 We have: x =t2, y = t3 - 3t dx/dt = 2t , dy/dt = 3t2 -3 dy/dx = (dy/dt) / (dx/dt) = (3t2 -3) / 2t The curve has horizontal tangent when dy/dt = 3t2 -3 =0, while dx/dt = 2t ≠ 0 → t = 1 or t = -1 Why? At t =1 →x=(1) 2=1 & y = (1)3– 3(1) = 1 - 3 = -2 At t =-1 →x=(-1) 2=1 & y = (-1)3 – 3(-1) = -1 + 3 = 2 Thus the curve has horizontal tangent at the points: (1,2) & (1,-2) What’s the equations of these tangents? Solution-Part 3 We have: x =t2, y = t3 - 3t dx/dt = 2t , dy/dt = 3t2 -3 dy/dx = (dy/dt) / (dx/dt) = (3t2 -3) / 2t The curve has vertical tangent when dy/dt = 3t2 -3 ≠ 0, while dx/dt = 2t = 0 → t = 0 Why? At t =0 →x=(0) 2=0 & y = (0)3– 3(0) = 0 Thus the curve has vertical tangent at the point (0,0) What’s the equations of this tangent? Solution-4** We have: dx/dt = 2t , dy/dt = 3t2 -3 dy/dx = (dy/dt) / (dx/dt) = (3t2 -3) / 2t = (3/2)t – (3/2)t-1 d2y/dx2 = d(dy/dx )/dx= [d(dy/dx)/dt] / [dx/dt] = [(3/2) +(3/2)t-2 ] / 2t = (3/4) t-1 + (3/4)t-3 = [3t2 + 3] / 4t3 d2y/dx2 > 0 if t > 0 & d2y/dx2 < 0 if t < 0 Thus the curve is concave upward if t > 0 and downward if t<0 We had: x =t2, y = t3 - 3t (t > 0 on the first quadrant. Why? and t < 0 on the fourth quadrant. Why? Book Example (2) Let x =r(t - sint), y =r(1 - cost), Where r is a constant 1. Find the slope of the tangent at t = π/3 2. Determine at which point (points), the graph has a horizontal tangent. 3. . Determine at which point (points), the graph has a vertical tangent Solution-Part1 We have: x =r(t - sint), y =r(1 - cost), dx/dt = r(1 - cost), , dy/dt = rsint → dy/dx = (dy/dt) / (dx/dt) = rsint / r(1 - cost) = sint / (1 - cost) (dy/dx)( π/3) = sin( π/3) / [1 - cos( π/3)] = (√3/2)/[1-(1/2)] = √3 Solution-Part 2 We have: x =r(t - sint), y =r(1 - cost), dx/dt = r(1 - cost), , dy/dt = rsint → dy/dx = (dy/dt) / (dx/dt) = rsint / r(1 - cost) = sint / (1 - cost) The curve has horizontal tangent when dy/dt = rsint =0, while dx/dt = r(1 - cost), ≠ 0 → t = nπ and t ≠ 2nπ Why? → t=(2n-1) π ; n is an integer. →x =r[(2n-1) π - 0)] = r(2n-1) π , y =r(1 – (-1) = 2r, Thus the curve has horizontal tangent at the points: (r(2n-1) π ,2r) Examples: ….,(-3rπ ,2r), (-rπ ,2r), (rπ ,2r), (3rπ ,2r), … Solution-Part 3 We have: x =r(t - sint), y =r(1 - cost), dx/dt = r(1 - cost), , dy/dt = rsint → dy/dx = (dy/dt) / (dx/dt) = rsint / r(1 - cost) = sint / (1 - cost) r(1 - cost), = 0 → t = 2nπ ; n is an integer Why? →x =r[(2nπ - 0)] = 2rnπ , y =r(1 – (1) = 0 Checking: show that dy/dx → + ∞ as t → 2nπ from the right* Thus the curve has vertical tangent at the points: (2nrπ , 0) Examples:…, (-4rπ , 0), (-2rπ , 0), (0, 0), (2rπ , 0), (4π , 0), …. Checking: show that dy/dx → + ∞ as t →2nπ from the right lim t 2 n dy dx lim t 2 n lim t 2 n lim t 2 n sin t 1 cos t cos t sin t cot( t )