Answer:
x = 1 - 5t
y = t
z = 1 - 5t
Step-by-step explanation:
For the equation of a line, we need a point and a direction vector. We are given a point (1, 0, 1).
Since the line is suppose to be a tangent to the given curve at the point (1, 0, 1), we need to find a tangent vector for which the curve passes through that point.
We have x = e^(-5t)cos5t
at t = 1, x = e^(-5)cos5
at t = 0, x = 1
y = e^(-5t)sin5t
at t = 1, y = e^(-5)sin5
at t = 0, y = 0
z = e^(-5t)
at t = 1, z = e^(-5)
at t = 0, z = 1
Clearly, the only parameter value for which the curve passes through the point (1, 0, 1) is t = 0.
In vector notation, the curve
r(t) = xi + yj + zk
= e^(-5t)cos5t i + e^(-5t)sin5t j + e^(-5t) k
r'(t) = [-5e^(-5t)cos5t - e^(-5t)sin5t] i +[e^(-5t)cos5t - 5e^(-5t)sin5t] j - 5e^(-5t) k
r'(0) = -5i + j - 5k
is a vector tangent at the point.
We get the parametric equation from this.
x = x(0) + tx'(0)
= 1 - 5t
y = y(0) + ty'(0)
= t
z = z(0) + tz'(0)
= 1 - 5t
Answer: x = -2 and 4
Step-by-step explanation:
x^2 - 2x - 8 = 0
(x - 4)(x + 2) = 0
<em>In decimal form, a = 114.5; b = 25.5</em>
We will write this as a system of equations, where 
and 
are the two angles.
We will use substitution to solve this system. We know what equals, so we plug that into the second equation.
Combine the like terms.
Add 
on both sides.
Divide both sides by 
.
Now just subtract that from 
.
Answer:
They all look right but i'm not sure about the last one?
Step-by-step explanation:
