HILLGROVE
Notes: Derivatives Of Parametric Equations PP&V Day 2Slope of the Tangent Line: Let Ct=(xt,yt) where x(t) and y(t) are differentiable. Assume that x'(t) is continuous and x'(t)≠0. Thendydx=dydtdxdt=y'(t)x'(t)Example One: Let ct=(t2+1,t3-4t). Find the equation of the tangent line at t=3 and find the points where the tangent is horizontal.3648075-99695d2ydx2=x'ty''t-y'tx''(t)x'(t)34000020000d2ydx2=x'ty''t-y'tx''(t)x'(t)3The Second Derivative: of a parametric equationExample Two: Find d2ydx2, given x=8t+9, y=1-4t, and t=-3 PP&V Day 2Remember:Derivativef(x) The original functionf'x>0f'x<0f''x>0f''x<0Try It: Given Ct=(-t2, 1-t24)195262511366500A. Graph the function and write in rectangular form.-1524005778500B. Find the tangent line at t=1C. Find the second derivative at t=-2 PP&V Day 2Example Three: A curve C is defined by the parametric equations x=t2, y=t3-3t114300057150t x=t2y=t3-3t-2-101200t x=t2y=t3-3t-2-1012396113059055400000A.) Graph CB.) Show that C has two tangents at the point (3,0) and find their equations.C.) Determine where the curve is concave upward or concave downward.D.) Determine if the graph has any vertical tangents or horizontal tangents and list the points where they are located.Proof of d2ydx2=x'ty''t-y'tx''(t)x'(t)3 ................
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