Math - Differentiation
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Hi i'm really stuck at this question under applications of differentiation. anyone able to help? thanks in advance
A curve C is given parametrically by the equations:
x = 2 + t , y = 1 - t^2
Show that the normal at the point with parameter t has equation:
x - 2ty = 2t^3 - t + 2 -
dx/dt = 1Originally posted by aeskay:Hi i'm really stuck at this question under applications of differentiation. anyone able to help? thanks in advance
A curve C is given parametrically by the equations:
x = 2 + t , y = 1 - t^2
Show that the normal at the point with parameter t has equation:
x - 2ty = 2t^3 - t + 2
dy/dt = -2t
hence dy/dx = dy/dt divide by dx/dt = -2t
Hence, gradient of tangent at any point = -2t
Which means that
gradient of normal at any point = 1/2t since the 2 gradients must multiply to -1
We can understand by at the point
(2+t, 1-t^2), gradient of normal = 1/2t
Hence, using equation of straight line y = mx + c
y = 1/2t x + c
to find c, sub point (2+t, 1-t^2)
1-t^2 = 1/2t (2+t) + c
c= 1/2 - t^2 - 1/t
Thus,
y = 1/2t x + 1/2 - t^2 - 1/t
Multiply by 2t throughout
2ty = x + t - 2t^3 - 2
Rearranging gives you
x - 2ty = 2t^3 -t + 2 -
!!! are you from cjc? i did this question from some additional questions my teacher gave us...Originally posted by aeskay:Hi i'm really stuck at this question under applications of differentiation. anyone able to help? thanks in advance
A curve C is given parametrically by the equations:
x = 2 + t , y = 1 - t^2
Show that the normal at the point with parameter t has equation:
x - 2ty = 2t^3 - t + 2 -
haha i think i'm getting it already.
nah i'm from nyjc. maybe the teachers frm different jcs share info too?
haha thanks for the help
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sians nid to dig out my amaths txtbook. getting rusty oreadi nxt year taking eg2.