Q:

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter. x = 4 + ln (t), y = t^2 + 6, (4, 7)

Accepted Solution

A:
Answer:y = 2x − 1Step-by-step explanation:By eliminating the parameter, first solve for t:x = 4 + ln(t)x − 4 = ln(t)e^(x − 4) = tSubstitute:y = t² + 6y = (e^(x − 4))² + 6y = e^(2x − 8) + 6Taking 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 = 2Writing equation of line using point-slope form:y − 7 = 2 (x − 4)y = 2x − 1Now, without eliminating the parameter, take derivative with respect to t:x = 4 + ln(t)dx/dt = 1/ty = t² + 6dy/dt = 2tFinding dy/dx:dy/dx = (dy/dt) / (dx/dt)dy/dx = (2t) / (1/t)dy/dx = 2t²At the point (4, 7), t = 1.  Evaluating the derivative:dy/dx = 2(1)²dy/dx = 2Writing equation of line using point-slope form:y − 7 = 2 (x − 4)y = 2x − 1