A function of two variables is given by, f(x,y) = e^2x-3y Find the tangent approximation to f(0.989,1.166) near (0,0), giving your answer to 4 decimal places.
Any kind soul to help on the problem. My professor raise this problem to me. i am so stuck.
any kind soul please help me. thanks.
Wow. The point is not very near (0,0,1), but
z-1 = ∂z/∂x (x-0) + ∂z/∂y (y-0)
z - 1 = 2x - 3y
z = 2x-3y+1
so plug in your (x,y) and chug away
a nice article is at
http://tutorial.math.lamar.edu/Classes/CalcIII/TangentPlanes.aspx
To find the tangent approximation to the function f(x, y) = e^(2x - 3y) near the point (0, 0), you can use partial derivatives and the equation of a tangent plane.
First, let's find the partial derivatives of f(x, y) with respect to x and y.
The partial derivative of f(x, y) with respect to x (denoted as ∂f/∂x) can be found by treating y as a constant and differentiating with respect to x. Since e^(2x - 3y) is an exponential function, the derivative will be the original function multiplied by the derivative of the exponent. In this case, the derivative of 2x with respect to x is 2. Thus, ∂f/∂x = 2e^(2x - 3y).
Similarly, to find the partial derivative of f(x, y) with respect to y (denoted as ∂f/∂y), we treat x as a constant and differentiate with respect to y. The derivative of -3y with respect to y is -3. Therefore, ∂f/∂y = -3e^(2x - 3y).
Now, we can use these partial derivatives to find the equation of the tangent plane at (0, 0). The equation of the tangent plane is given by:
z - f(0,0) = (∂f/∂x)(x - 0) + (∂f/∂y)(y - 0)
Since we are interested in finding the tangent approximation at (0.989, 1.166), we substitute these values into the equation of the tangent plane:
z - f(0,0) = (∂f/∂x)(0.989 - 0) + (∂f/∂y)(1.166 - 0)
Now, we can calculate the values for ∂f/∂x and ∂f/∂y at (0, 0):
∂f/∂x = 2e^(2(0) - 3(0)) = 2e^0 = 2
∂f/∂y = -3e^(2(0) - 3(0)) = -3e^0 = -3
Plugging these values into the equation of the tangent plane:
z - f(0,0) = (2)(0.989 - 0) + (-3)(1.166 - 0)
Now, evaluate the right-hand side of the equation:
z - f(0,0) = 1.978 - 3.498
To find the tangent approximation at (0.989, 1.166), we need to find z. Rearranging the equation:
z = f(0,0) + 1.978 - 3.498
Now, calculate f(0,0) = e^(2(0) - 3(0)) = e^0 = 1:
z = 1 + 1.978 - 3.498
z = -0.52
Therefore, the tangent approximation to f at (0.989, 1.166) near (0, 0) is z = -0.52.
Note: The answer may differ depending on the precision used during calculations, so be sure to use the given precision requirement of 4 decimal places.