What is the answer to r(t)=(3-3t)i+4tj,0<= t<= 1 ?

The answer to r(t)=(3-3t)i+4tj,0<= t<= 1 is r=4j,t=1

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r(t)=(3-3t)i+4tj,0<= t<= 1

r (t) = (3 - 3 t) i + 4 t j, 0 \leq t \leq 1

r = 4 j, t = 1

Solution steps

r (t) = (3 - 3 t) i + 4 t j

Rewrite

r (t)

in standard complex form:

r (t) + 0 i

r (t) + 0 i = (3 - 3 t) i + 4 t j

Rewrite

(3 - 3 t) i + 4 t j

in standard complex form:

4 j t + (3 - 3 t) i

r (t) + 0 i = 4 j t + (3 - 3 t) i

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

[r (t) = 4 j t 0 = 3 - 3 t]

[r (t) = 4 j t 0 = 3 - 3 t] : r = 4 j, t = 1

r = 4 j, t = 1

\frac{a - 6 i}{- 2 + 3 i} = b - 7

x, x^{2} + 2 x i - 5 = 3 i

i (x) = \frac{2 x + 1}{x - 1}

∣ z - 2 i ∣ = 4

1 + 3 i e z^{2} = (1 - i)

What is the answer to r(t)=(3-3t)i+4tj,0<= t<= 1 ?
The answer to r(t)=(3-3t)i+4tj,0<= t<= 1 is r=4j,t=1