Respuesta :
Answer:
y = 0
y =2x
Step-by-step explanation:
Given parametric equations:
x (t) = sin (t)
y (t) = sin (t + sin (t))
The slope of the curve at any given point is given by dy / dx we will use chain rule to find dy / dx
(dy / dx) * (dx / dt) = (dy / dt)
(dy / dx) = (dy / dt) / (dx / dt)
Evaluate dx / dt and dy / dt
dx / dt = cos (t)
dy / dt = cos (t + sin (t)) * (1+cos (t))
Hence,
dy / dx = (1+cos(t))*cos(t + sin (t))) / cos (t)
@Given point (x,y) = 0 we evaluate t
0 = sin (t)
t = 0 , pi
Input two values of t and compute dy / dx
@ t = 0
dy / dx = (1 + cos (0))*cos (0 + sin (0))) / cos (0)
dy / dx = (1+1)*(1) / (1) = 2 @ t = 0
@t = pi
dy / dx = ( 1 + cos (pi))* cos (pi + sin (pi)) / cos (pi)
dy / dx = (1-1) * (-1) / (-1) = 0 @ t = pi
The corresponding gradients are 0 and 2 in increasing order and their respective equations are:
y = 2x
y = 0
The equation of the two tangent lines at the point (x,y) = (0,0) in order of increasing slope are; y = 0 and y = 2x
We are given the parametric equations of the curve as;
x = sin(t)
y = sin(t + sin(t))
Now, since we want to find slope, the we need to find dy/dx from;
dy/dx = (dy/dt) ÷ (dx/dt)
Thus;
dx/dt = cos(t)
Using chain rule;
dy/dt = cos (t + sin(t)) × (1 + cos(t))
Thus;
dy/dx = [cos (t + sin(t)) × (1 + cos(t))]/(cos(t))
At (0, 0), we have;
0 = sin(t) ---(1)
0 = sin(t + sin(t)) ---(2)
From eq (1), values of t that makes the function 0 are;
0 and π
Thus;
At t = 0;
dy/dx = [cos (0 + sin(0)) × (1 + cos(0))]/(cos(0))
dy/dx = (1 + 1)/1
dy/dx = 2
At t = π;
dy/dx = [cos (π + sin(π)) × (1 + cos(π))]/(cos(π))
dy/dx = 0
Using the point slope form; y - y₁ = m(x - x₁)
At m = 2, we have;
y - 0 = 2(x - 0)
y = 2x
At m = 0, we have;
y - 0 = 0(x - 0)
y = 0
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