A function of 2 variables is given by,
f(x,y) = exp(2x-3y)
Find the tangent approximation to
f(1.155,0.632) near (0,0).
Thanks!
To find the tangent approximation to a function at a specific point, we can use the linearization. The tangent approximation is given by:
f(x, y) ≈ f(a, b) + (∂f/∂x)(x − a) + (∂f/∂y)(y − b)
Where (a, b) is the point near which we want to approximate the function and (∂f/∂x) and (∂f/∂y) are the partial derivatives of the function.
Let's find the tangent approximation to f(1.155, 0.632) near (0, 0) using the given function f(x, y) = exp(2x - 3y).
Step 1: Calculate (∂f/∂x) and (∂f/∂y).
∂f/∂x = d/dx (exp(2x - 3y)) = 2exp(2x - 3y)
∂f/∂y = d/dy (exp(2x - 3y)) = -3exp(2x - 3y)
Step 2: Plug in the values into the tangent approximation formula.
f(1.155, 0.632) ≈ f(0, 0) + (∂f/∂x)(1.155 - 0) + (∂f/∂y)(0.632 - 0)
Step 3: Calculate the values.
Using the values of (∂f/∂x) = 2exp(2x - 3y) and (∂f/∂y) = -3exp(2x - 3y) at the point (0, 0):
f(1.155, 0.632) ≈ exp(0) + 2exp(0) * (1.155 - 0) + (-3exp(0)) * (0.632 - 0)
Simplifying the expression:
f(1.155, 0.632) ≈ 1 + 2 * 1.155 - 3 * 0.632
f(1.155, 0.632) ≈ 1 + 2.31 - 1.896
f(1.155, 0.632) ≈ 1.414
Therefore, the tangent approximation to f(1.155, 0.632) near (0, 0) is approximately 1.414.
Please note that this is an approximation and may not be as accurate as the actual value of f(1.155, 0.632). The tangent approximation becomes more accurate as the point of approximation gets closer to the given point.