What is the answer to r(t)=(2t+1)i+(t^2-1)j ?

The answer to r(t)=(2t+1)i+(t^2-1)j is

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r(t)=(2t+1)i+(t^2-1)j

r (t) = (2 t + 1) i + (t^{2} - 1) j

(r = \frac{3}{2} j, t = - \frac{1}{2})

Solution steps

r (t) = (2 t + 1) i + (t^{2} - 1) j

Rewrite

r (t)

in standard complex form:

r (t) + 0 i

r (t) + 0 i = (2 t + 1) i + (t^{2} - 1) j

Rewrite

(2 t + 1) i + (t^{2} - 1) j

in standard complex form:

(j t^{2} - j) + (2 t + 1) i

r (t) + 0 i = (j t^{2} - j) + (2 t + 1) i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[r (t) = j t^{2} - j 0 = 2 t + 1]

[r (t) = j t^{2} - j 0 = 2 t + 1] : (r = \frac{3}{2} j, t = - \frac{1}{2})

(r = \frac{3}{2} j, t = - \frac{1}{2})

2 x + 51 = 4 (2 + y i) - 12

z^{2} - (1 + 9 i) z - 20 + 5 i = 0

z = z (a + i)

\frac{x + 1 + i ( y - 3 )}{5 + 3 i} = 1 + i

v \cdot v = i - 5 j

What is the answer to r(t)=(2t+1)i+(t^2-1)j ?
The answer to r(t)=(2t+1)i+(t^2-1)j is