Question

Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter.

x = e^sqrt(t)
y = t - ln t2
t = 1
y(x) =

Answer 1

y(x) = -(2/e)x +3

Explanation:

You want the equation of the line tangent to the parametric curve at t=1.

(x, y) = (e^(√t), t -2·ln(t))

Point

At t=1, the point of tangency is ...

(x, y) = (e^(√1), 1 -2·ln(1)) = (e, 1)

Slope

The derivatives with respect to t are found using the chain rule:

dx = d(e^u)du = d(e^√t)(1/(2√t))dt

dx = (e^√t)/(2√t))·dt

dy = (1 -2/t)·dt

Then the slope of the tangent line is ...

m = dy/dx = (1 -2/t)(2√t)/e^√t

For t=1, this is ...

m = (1 -2/1)(2√1)/(e^1) = -2/e

Point-slope equation

The equation for a line with slope m through point (h, k) is ...

y = m(x -h) +k

The equation for a line with slope -2/e through point (e, 1) is ...

y = (-2/e)(x -e) +1

y = (-2/e)x +3