Consider the given vector equation. r(t) = 2eti + 3e−tj (a) find r'(t). r'(t) = <2et,−3e−t> (b) sketch the plane curve tog
ether with the position vector r(t) and the tangent vector r'(t) for the given value of t = 0.
1 answer:
(a) Differentiate each of the components to get r'(t). The rule is
.. d/dt (a*e^(bt)) = a*b*e^(bt)
The answer you have shown is the correct one.
(b) See the figure. The red curve is the position r(t) for 0 ≤ t ≤ 2. The dashed orange line is the tangent line, whose equation is
.. L(t) = r(0) +r'(0)*t = (2 +2t)i +(3 -3t)j
.. d/dt (a*e^(bt)) = a*b*e^(bt)
The answer you have shown is the correct one.
(b) See the figure. The red curve is the position r(t) for 0 ≤ t ≤ 2. The dashed orange line is the tangent line, whose equation is
.. L(t) = r(0) +r'(0)*t = (2 +2t)i +(3 -3t)j
You might be interested in
Notice that
, so
Then taking the positive square root gives
so and
.