A curve has parametric equations: x=(t-1)^3 and y= 3t - 8/(t^2). Find dy/dx in terms of t. Then find the equation of the normal at the point on the curve where t=2.
dx/dt = 3(t-1)2dy/dt = 3 + 16t-3dy/dx=(dy/dt)(dt/dx) dy/dx = 3 + 16t-3 / 3(t-1)2
At t=2 dy/dx= (3 + 16/8) / 3 = 5/3 Gradient of the normal = -3/5with t=2 y-4=0x-1=0 y=mx + c y - 4 = -3/5(x-1) 3x +5y -23 = 0
JH
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