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Fed [463]
1 year ago
7

The slope of the line below is 3. Which of the following is the point slopeform of the ling?O A. y-2=3(446)O B. y-6--3/442)O C.

y425-34-5)O D. 446=-3x - 2)

Mathematics
1 answer:
polet [3.4K]1 year ago
4 0

The slope of the line is given as -3 and it passes through the point (2, -6).

Thus we can write the equation of the line in point-slope form;

\begin{gathered} \frac{y-y_1}{x-x_1}=-3 \\ \frac{y-(-6)}{x-2}=-3 \\ \frac{y+6}{x-2}=-3 \\ y+6=-3(x-2) \\  \end{gathered}

Therefore, the correct option is D

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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

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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

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Answer:

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