Solution:
B(t) = ∑ki=0 Pk Bk,n(t) where, 0<=t<=1
It is given that curve passes (0,0) and (-4,2), means starting point of the curve is P0 = (0,0) and P3 = (-4,2)
Whereas the curve is controlled by P1 (14,10) and P2 (4,0)
Number of control points,
P0=(0,0) , P1=(14,10) , P2=(4,0) , P3=(-4,2)
Degree of equation n =Control_point -1 = 4 -1 = 3
B(t) = P0B0,3(t)+ P1B1,3(t) + P2B2,3(t)+ P3B3,3(t)
Where, Bk,n(t)= nCk uk (1-t)n-k
B0,3(t)= 3C0 t0 (1-t)3-0
= 1(1-t) 3
= 1(1-t) 3
= (1-t)3
B1,3(t)= 3C1 u1 (1-t)3-1
= t(1-t) 2
= t(1-t) 2
= 3t (1-t)2
B2,3(t)= 3C2 t2 (1-t)3-2
= t2(1-t) 1
= t2(1-t) 1
= 3t2 (1-t)1
B3,3(t)= 3C3 u3 (1-t)3-3
= t3(1-t) 0
= t3(1-t) 0
= t3
B(t)= P0 (1-t)3 + P1 3t (1-t)2+ P2 3t2 (1-t)1+ P3 t3
Now, Using this equation lets calculate equation for x control point
P0=(0,0) , P1=(14,10) , P2=(4,0) , P3=(-4,2)
= 0(1-t)3 + (14) 3t (1-t)2+ (4) 3t2 (1-t)+ (-4) t3
= 42t(t2+1-2t)+ 12t2(1-t)-4t3
= 42t3+42t-84t2+12t2-12t3-4t3
=26t3-72t2+42t
Using this equation lets calculate equation for y control points
= 0(1-t)3 + (10) 3t (1-t)2+ (0) 3t2 (1-t)+ (2) t3
=30t (1+t2-2t)+2t3
=30t+30t3-60t2+2t3
= 32t3 -60t2+30t
So, The answer is P(t) =[ x(t) y(t)] =[ (26t3-72t2+42t) (32t3 -60t2+30t)]
