x^3 + xtan(y) = 27 passes through (3,0) Use linearization to estimate the value of y at x=3.1.. I have no idea what linearization even is.. Can someone explain?
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local linearization is the same thing as euler's method of approximation.
essentially your creating tiny little line segments branching off your initial point and subsequent estimated points using the derivative.
the derivative in this case is 3x^2 + tan(y), so the slope of your first tiny line segment is 3(3)^2 + tan(0) which is equal to 27.
usually you'd have to introduce step sizes into the process because the desired x value is usually more than a tenth away from the initial, but here we're just going to multiply the slope by the difference in x value, 0.1: (0.1)(27)= 2.7
now we add that too our initial y value, 0 in this case, and get out 2.7
thats is your approximate y value at 3.1
hope i helped!!
essentially your creating tiny little line segments branching off your initial point and subsequent estimated points using the derivative.
the derivative in this case is 3x^2 + tan(y), so the slope of your first tiny line segment is 3(3)^2 + tan(0) which is equal to 27.
usually you'd have to introduce step sizes into the process because the desired x value is usually more than a tenth away from the initial, but here we're just going to multiply the slope by the difference in x value, 0.1: (0.1)(27)= 2.7
now we add that too our initial y value, 0 in this case, and get out 2.7
thats is your approximate y value at 3.1
hope i helped!!