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Maths problem!!! Pls Help!
發問:
Conic in R2 given by the parametrization P(t) = (t-1, t2 – 2t) (i)Show that the conic is a parabaola with equation x2 = y +1 (ii)Find the gradient of the tangent to this parabola at the point with parameter t, for t≠1 (iii)Find the equation of the tangent to the parabola at the point P(0) and P(2).
(i) x=t-1, y=t^2-2t y=(t^2-2t+1)-1 y=[(t-1)^2]-1 y=x^2-1----------(1) x^2=y+1 (ii) x^2=y+1 y=x^2-1 dy/dx=d(x^2-1)/dx dy/dx=2x----------(2) (dy/dt)/(dx/dt)=2(t-1)------(3) gradient of the tangent to this parabola at the point with parameter t =2(t-1) (iii) it is easy and direct to use the equations (1) and (2) with variables x,y if you use equation (3), you need to responding value of t for finding y and dy/dx. more steps to be needed for t but you can it for giving the same answer. P(0) means x=0, substitute x=0 into (1), y=[(0)^2] -1= -1 substitute x=0 into (2), dy/dx=2(0)=0 the equation of the tangent to the parabola at the point P(0) is [y- (-1)]/[(x- (0)]=0 (y+1)/x=0 y+1=0 or y= -1 P(2) means x=2, substitute x=2 into (1), y=[(2)^2] -1=3 substitute x=2 into (2), dy/dx=2(2)=4 the equation of the tangent to the parabola at the point P(2) is [y- (3)]/[(x- (2)]=4 y -3=4(x -2) y -3=4x -8 y=4x -5 or 4x -y -5=0
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Maths problem!!! Pls Help!
發問:
Conic in R2 given by the parametrization P(t) = (t-1, t2 – 2t) (i)Show that the conic is a parabaola with equation x2 = y +1 (ii)Find the gradient of the tangent to this parabola at the point with parameter t, for t≠1 (iii)Find the equation of the tangent to the parabola at the point P(0) and P(2).
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最佳解答:(i) x=t-1, y=t^2-2t y=(t^2-2t+1)-1 y=[(t-1)^2]-1 y=x^2-1----------(1) x^2=y+1 (ii) x^2=y+1 y=x^2-1 dy/dx=d(x^2-1)/dx dy/dx=2x----------(2) (dy/dt)/(dx/dt)=2(t-1)------(3) gradient of the tangent to this parabola at the point with parameter t =2(t-1) (iii) it is easy and direct to use the equations (1) and (2) with variables x,y if you use equation (3), you need to responding value of t for finding y and dy/dx. more steps to be needed for t but you can it for giving the same answer. P(0) means x=0, substitute x=0 into (1), y=[(0)^2] -1= -1 substitute x=0 into (2), dy/dx=2(0)=0 the equation of the tangent to the parabola at the point P(0) is [y- (-1)]/[(x- (0)]=0 (y+1)/x=0 y+1=0 or y= -1 P(2) means x=2, substitute x=2 into (1), y=[(2)^2] -1=3 substitute x=2 into (2), dy/dx=2(2)=4 the equation of the tangent to the parabola at the point P(2) is [y- (3)]/[(x- (2)]=4 y -3=4(x -2) y -3=4x -8 y=4x -5 or 4x -y -5=0
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