Find An Equation Of The Tangent To The Curve At The Point Corresponding To The Given Value Of The Parameter X Evty T Ln 1 (20.41 KiB) Viewed 34 times
Find An Equation Of The Tangent To The Curve At The Point Corresponding To The Given Value Of The Parameter X Evty T Ln 2 (25.69 KiB) Viewed 34 times
Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. x=evty=t−ln(t3)t=1 When a curve is given by the parametric equations x=nt) and y=g(t), then dxdy can be found by caloulating dxdy=drdydy=f(t)gy(c) We Nave x=2t and y=t−1 a t5. Since x=f(t)=kt is a composilion, then by the Chain hule we have
Ise the given parameters to answer the following questions. If you have a graphing device, graph the curve to check your work. x=2t3+3t2−180ty=2t3+3t2+5 (a) Find the points on the curve where the tangent is horizontal. )( smaller t) ) (larger t ) (b) Find the points on the curve where the tangent is vertical. ) (smaller t) ) (larger t )
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