Question: Let x(t) = t' - 4t' + 1 and y(t) = t - 5t- + 2t + 8 At t = 1, 7- (1) =

 Let x(t) = t' - 4t' + 1 and y(t) =t - 5t- + 2t + 8 At t = 1, 7-(1) = y(1) -7. dt 1=1= dy dy It=1 = tangent slope= speed (1)\fLet x(t) = t - sin(t) CYCLOID and y(t) =1 - cos(t) - At t = 1, r(1) = 2 35 6 7 y(1) da dt It=1= dy at =1= dy It=1= tangent slope = speed (1) =Let x(t) = 5 cos(t) ELLIPSEand y(t) = 3 sin(t) At t = 5, T(5) = -2y(5) = dt dy dt t=5= dy It= = tangent slope speedimage text in transcribedimage text in transcribedimage text in transcribed

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Let x(t) = t' - 4t' + 1 and y(t) = t - 5t- + 2t + 8 At t = 1, 7- (1) = y(1) -7. dt 1=1= dy dy It=1 = tangent slope = speed (1)\fLet x(t) = t - sin(t) CYCLOID and y(t) = 1 - cos(t) - At t = 1, r(1) = 2 3 5 6 7 y(1) da dt It=1= dy at =1= dy It=1 = tangent slope = speed (1) =Let x(t) = 5 cos(t) ELLIPSE and y(t) = 3 sin(t) At t = 5, T(5) = -2 y(5) = dt dy dt t=5= dy It= = tangent slope speed

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