Find the parametric equations for the line that is tangent to the given curve at the given parameter (3cost)i (t^4 -3sint)j

Mathematics

1 answer:

ale4655 [162]3 years ago

5 0

Answer:

Following are the solution to this question:

Step-by-step explanation:

Please find the complete question in the attached file.

In the given equation, when the point t=0

So,

$\to r(0) = (3 \cos 0)i + (0^4 - 6 \sin 0)j + (2e^{3\times 0})k)$

$= (3 \times 1)i + (0 - 0)j + (2e^{0})k)\\\\ = 3i + 0j + (2 \times 1)k)\\\\ = 3i + 0j + 2k \\$

The value of the coordinates are $3, 0, 2$ . so, the equation of the line is:

$\to \frac{(x-3)}{3 \cos \ t} = \frac{(y-0)}{(t^{4}-6 \sin \ t)} = \frac{(z-2)}{2e^{3t}}=k$

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I’m confused and don’t have an angle tool to even try to do the question

yanalaym [24]

<h3>Answer: 40</h3>

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Explanation:

JQ is longer than QN. We can see this visually, but the rule for something like this is the segment from the vertex to the centroid is longer compared to the segment that spans from the centroid to the midpoint.

See the diagram below.

The ratio of these two lengths is 2:1, meaning that JQ is twice as long compared to QN. This is one property of the segments that form when we construct the centroid (recall that the centroid is the intersection of the medians)

We know that JN = 60

Let x = JQ and y = QN

The ratio of x to y is x/y and this is 2/1

x/y = 2/1

1*x = y*2

x = 2y

Now use the segment addition postulate

JQ + QN = JN

x + y = 60

2y + y = 60

3y = 60

y = 60/3