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Question
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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#### randy patton

##### Expertise

college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography

##### Experience

26 years as a professional scientist conducting academic quality research on mostly classified projects involving math/physics modeling and simulation, data analysis and signal processing, instrument development; often ocean related

Publications
J. Physical Oceanography, 1984 "A Numerical Model for Low-Frequency Equatorial Dynamics", with M. Cane

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M.S. MIT Physical Oceanography, B.S. UC Berkeley Applied Math

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Also an Expert in Oceanography