SEFY workshop

 Implicit and Parametric DifferentiationProblem set Calculate dydx for each of the followingy2=x3-x+1x=2t, y=3t2-2y=sinxcos2yx= csct, y=2t3-lnt3x=y-2xyA curve has equation 3x2+y2-xy=4. The points P and Q lie on the curve. The gradient of the curve is 85 at both points P and Q.Show using implicit differentiation that y+2x=0 at P and Q.Find the coordinates of P and Q.Consider the curve with parametric equation x=sec2θ, y=cotθ, -π2≤θ≤π2.Calculate dydx, expressing your answer in terms of x and y.Calculate a Cartesian equation for the curve, and check your answer to part a. by differentiating implicitly.The point P on the curve occurs when θ=π4. Calculate the equation of the tangent at P, expressing your answer in the form ax+by=c, where a, b and c are integers.The tangent at P crosses the curve at one other point, Q. Calculate the coordinates of P and Q, and hence show that the line segment PQ has length 325. Find and classify the stationary points of the curve with implicit equation 2-ye3x+3x=y2.Find a parameterization that traces the ellipse (x-2)29+(y+3)24=1,starting at the point (-1,-3) in an anticlockwise direction.A cycloid is a curve that describes the motion of a fixed point P on the circumference of a circle as it rolls along the x-axis at a constant speed.The position of the point P(t)=(x(t),y(t)) at time t may be described by parametric equations 1066800173286x(t)=a(vt-sin(vt)), y(t)=a(1-cos(vt)), t≥0.Show that the position of the centre of the circle at time t is C(t)=(avt,a). What might be a good interpretation for the constants a and v?By considering stationary points, or otherwise, sketch the curve traced out by the point P.Show that dydx=2a-yy. This equation can be used to derive a Cartesian equation for the cycloid - soon you will have learnt enough integral calculus to do this!Answers dydx=3x2-12ydydx=-3t3(or dydx=24x3)dydx=cosxcos2y-2sinxsin2ydydx=(1t-6t2)tantsintdydx=3xln3+2y1-2x Differentiating 3x2+y2-xy=4 implicitly, we have 6x+2ydydx-y-xdydx=0. At P and Q, dydx=85, and we can substitute this in to get 6x+2y?85-y-85x=0, which simplifies to give 2x+y=0.The coordinates of P and Q may be found by solving the equations 3x2+y2-xy=4 and 2x+y=0 simultaneously. P and Q are equal to (23,-43) and (-23,43), in some order. dydx=-12y3y2=1x-1is one possibility.Substituting θ=π4 into the parametric equation for the curve gives P=(2,1), Therefore, at P, dydx=-12. The tangent at P has equation x+2y=4.Now we solve y2=1x-1 and x+2y=4 simultaneously, to get Q=(5,-12). The distance between P and Q is |PQ|=32+(32)2=325. Differentiating implicitly gives -dydxe3x-3ye3x+3=2ydydx, and hence dydx=3-3ye3x2y+e3x. Stationary points therefore occur when y=e-3x. Substituting this into the equation for the curve, we have 1+3x=e-6x. The curves y=1+3x and y=e-6x have only one point of intersection, namely when x=0. Hence the curve 2-ye3x+3x=y2 has only one stationary point, occurring when x=0. The corresponding y-value is y=1.1978025276225To classify this stationary point, recall that dydx=3-3ye3x2y+e3x, and calculate the second derivative, using the quotient rule:d2ydx2=(2y+e3x)(-3dydxe3x-9ye3x)-(3-3ye3x)(2dydx+3e3x)(2y+e3x)2. This looks horrible, but we don’t actually need to simplify it at all - we can just substitute the values x=0, y=1and dydx=0 straight in, to evaluate d2ydx2 at the point (0 ,1). We getd2ydx2=(2+1)(-9)-(3-3)(3)(2+1)2=-3<0,and hence the point (0,1) is a local maximum.The equation (x-2)29+(y+3)24=1 suggests we might try putting x-23=cosθ and y+32=sinθ. That is, x=3cosθ+2 and y=2sinθ-3. Starting at θ=-π and increasing to θ=+πensures the parameterization begins at (-1,-3) and moves anticlockwise. Given x(t)=a(vt-sin(vt)) and y(t)=a(1-cos(vt)), the squared-distance from P(t)=(x(t),y(t)) to the point (avt,a) is(x(t)-avt)2+(y(t)-a)2=a2sin2(vt)+a2cos2(vt) = a2.Hence at any given time t, P(t)lies on the circle with centre (avt,a) and radius a. The constant a represents the radius of the rolled circle , and v represents the velocity of the motion.The curve crosses the x-axis at the points where y(t)=0, i.e. when cos(vt)=1. That is, at t=2πnv, where n∈?. At these points, x(t)=2nπa, for n∈?. To calculate stationary points, note that dydx=x'(t)y'(t)=avsin(vt)av-avcos(vt)=sin(vt)1-cos(vt), and so dydx=0 whensin(vt)=0, i.e. when t=nπv, for some n∈?. However, we also require the denominator of this expression fordydx to be nonzero, that is cos(vt)≠1. As a result, stationary points occur when t=nπv and n is odd. Writing n=2k+1, the corresponding Cartesian coordinates are ((2k+1)πa,2a). Either by considering the physical interpretation, or by calculating the second derivative, one can check that each of these points is a local maximum.Note also that the curve is periodic with period 2πa. This can be seen either by physical intuition, or by substituting t=t0+2πnv into the parametric equations for the curve. 1202690142875See also: have already calculated dydx=sin(vt)1-cos(vt), and hence1+(dydx)2=1+sin2(vt)1-2cos(vt)+cos2(vt)=1-2cos(vt)+cos2+sin2(vt)1-2cos(vt)+cos2(vt)=2-2cos(vt)(1-cos(vt))2=21-cos(vt)=2ay.Rearranging, we have dydx=2ay-1=2a-yy.The Cartesian Equation of a CycloidFor anyone interested in calculating the Cartesian equation, it is easiest to invert this last equation and integratedxdy=y2a-ywith respect to y, to obtain an expression for x in terms of y. The MAS003 course does not expect you to be able to do this yet, but it’s there for anyone seeking a challenge! [Hint: try applying the substitution y=2asin2u.] ................
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