Question

x=3t^2+1且y=2t^3+1,在x=4,y=3的切线方程

Answer 1

令y=2t^3+1=3,∴t=1

dx/dt=6t,dy/dt=6t^2

∴dy/dx=(dy/dt)/(dx/dt)=t

所以切线的斜率k=t=1

∴切线方程为y-3=x-4,即x-y-1=0

望采纳