1. The graph of the parametric equations x= 2-3t, t= -1+5t is a line. What is the slope of this line?
Please help!!!!!
t = -1 + 5 t ? I doubt it
then
1 = 4 t
and
t = 1/4
x = 2 - 3 (1/4)
period
Probably meant to type:
x= 2-3t, y= -1+5t
from x= 2-3t, -----> t = (2-x)/3
rom y= -1+5t ---> t = (y+1)/5
(2-x)/3 = (y+1)/5
3y+3 = 10 - 5x
5x + 3y = 7
slope = -5/3
Yeah sorry I don't know why it keeps happening.
Why did the graph go to therapy? Because it had trouble finding its slope! Okay, let's get serious. To find the slope of this line, let's first note that both x and t are linear functions of t. The x-coordinate is given by x = 2 - 3t, and the y-coordinate is given by t = -1 + 5t. If we solve for t in terms of x, we get t = (2 - x) / 3. Now we have a line in slope-intercept form, y = mx + b, where m is the slope. Since y = t, we substitute in t = (2 - x) / 3, giving us y = (2 - x) / 3. The slope of this line is the coefficient of x, which is -1/3. So, the slope of the line is -1/3.
To find the slope of the line represented by the given parametric equations, we need to find the derivative of the x-coordinate with respect to the parameter t.
Let's start by differentiating the x-coordinate equation:
x = 2 - 3t
Taking the derivative of both sides with respect to t, we get:
dx/dt = -3
This derivative gives us the rate of change of x with respect to t, which is equivalent to the slope of the line. Therefore, the slope of the line represented by the given parametric equations is -3.
In summary, to find the slope of a line represented by parametric equations, we differentiate the x-coordinate equation with respect to the parameter.