Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

Describe (in words) the curves traced by particles moving in the xy-plane according to the parametric equations in (a–h). You do not have to draw the graphs yourself, but Desmos is still extremely helpful here (5t + 3, 2t − 6) with −2 ≤ t ≤ 3

Sagot :

Answer:

here we have:

x = 5*t + 3

y = 2*t - 6

both for -2 ≤ t ≤ 3

First, we can rewrite:

x = a*y + b

let's try to find the values of a and b.

5*t + 3 = a*(2*t - 6) + b

5*t + 3 = a*2*t + (b - 6*a)

Here we have two equations:

5 = a*2

3 = (b - 6*a)

from the first one, we can solve:

5/2 = a

replacing this into the second, we et:

3 = b - 6*(5/2)

3 = b - 15

3 + 15 = b = 18

Then:

x = (5/2)*y + 18

y =  2*t - 6

So, y is a linear equation that depends on the variable t

x is a linear equation that only depends on x (from this we already know that the graph will be a line)

and we know that  −2 ≤ t ≤ 3

Then the minimum value of y is:

y = 2*(-2) - 6 = -10

and the largest value of y is:

y = 2*3 - 6 = 0

So y-varies in the range between -10 and 0

And because x =  (5/2)*y + 18

The minimum value of x is:

x = (5/2)*-10 + 18 = -25 + 18 = -7

and the maximum value of x is:

x = (5/2)*0 + 18 = 18

Then we know that this line starts in the point (-10, -7) and ends in the point (0, 18)