Answer:
y = 2x − 1
Step-by-step explanation:
By eliminating the parameter, first solve for t:
x = 4 + ln(t)
x − 4 = ln(t)
e^(x − 4) = t
Substitute:
y = t² + 6
y = (e^(x − 4))² + 6
y = e^(2x − 8) + 6
Taking derivative using chain rule:
dy/dx = e^(2x − 8) (2)
dy/dx = 2 e^(2x − 8)
Evaluating at x = 4:
dy/dx = 2 e^(8 − 8)
dy/dx = 2
Writing equation of line using point-slope form:
y − 7 = 2 (x − 4)
y = 2x − 1
Now, without eliminating the parameter, take derivative with respect to t:
x = 4 + ln(t)
dx/dt = 1/t
y = t² + 6
dy/dt = 2t
Finding dy/dx:
dy/dx = (dy/dt) / (dx/dt)
dy/dx = (2t) / (1/t)
dy/dx = 2t²
At the point (4, 7), t = 1. Evaluating the derivative:
dy/dx = 2(1)²
dy/dx = 2
Writing equation of line using point-slope form:
y − 7 = 2 (x − 4)
y = 2x − 1
