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Find the equation of the line tangent to the curve at the given point: x=sqrt(t); y=t+1 at t=1

The parametric equation can be written as y = f(x) using t = x^2

y = x^2 + 1.

The eqn for a line is y = mx + b, where m is the slope dy/dx (evaluated at t = 1 in this case) and b the intercept b = f(0)

dy/dx = 2x = 2(t^1/2) = 2 at t = 1 and b = 1 for x = t = 0.

y = 2x+1.

The slope could also be obtained using the chain rule (with t = x^2) as dy/dx = (dy/dt)(dt/dx) = 1･(2x) = 2x.

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