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loris [4]
3 years ago
6

Line m and point P are shown below. Part A: Using a compass and straightedge, construct line n parallel to line m and passing th

rough point P. Leave all construction marks. Part B: Explain the process that you used to construct line n.

Mathematics
1 answer:
Morgarella [4.7K]3 years ago
8 0

Answer:

The steps to construct a a line parallel to another line from a point includes

1) From the given line draw a transversal through the point

2) With the compass, copy the angle formed between the transversal and the given line to the point P

3) Draw a line through the intersection of the arcs of the angle construction to get the parallel line through the point P

Step-by-step explanation:

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Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = e−5t
Elan Coil [88]

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

8 0
3 years ago
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