Example Question - using differentials
Here are examples of questions we've helped users solve.
Approximating Change using Differentials
To approximate the change in \( f(x, y, z) = 3x^2 - yz + 2 \) when \( (x, y, z) \) changes from \( (1, 2, 3) \) to \( (0.99, 2.01, 3.01) \), we will use differentials. First, we need to compute the partial derivatives of \( f \) with respect to \( x \), \( y \), and \( z \): \[ \frac{\partial f}{\partial x} = 6x \] \[ \frac{\partial f}{\partial y} = -z \] \[ \frac{\partial f}{\partial z} = -y \] Now we evaluate these partial derivatives at the point \( (1, 2, 3) \): \[ \frac{\partial f}{\partial x}(1, 2, 3) = 6(1) = 6 \] \[ \frac{\partial f}{\partial y}(1, 2, 3) = -(3) = -3 \] \[ \frac{\partial f}{\partial z}(1, 2, 3) = -(2) = -2 \] The differential approximates the change in \( f \) as follows: \[ df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz \] Since \( dx = 0.99 - 1 = -0.01 \), \( dy = 2.01 - 2 = 0.01 \), and \( dz = 3.01 - 3 = 0.01 \), we can substitute these values into the equation to find \( df \): \[ df = (6)(-0.01) + (-3)(0.01) + (-2)(0.01) \] \[ df = -0.06 - 0.03 - 0.02 \] \[ df = -0.11 \] Thus, the differential \( df \) approximates the change in \( f \) when \( (x, y, z) \) changes from \( (1, 2, 3) \) to \( (0.99, 2.01, 3.01) \) to be approximately \( -0.11 \).
Approximating Temperature Difference using Differentials
The problem asks us to use differentials to approximate the temperature difference between two points in an \( xyz \) coordinate system where the temperature \( T \) is given by the function \( T = xy + yz + zx \). To approximate the change in \( T \) when moving from point \( (x_0, y_0, z_0) \) to \( (x_0 + \Delta x, y_0 + \Delta y, z_0 + \Delta z) \), we can use differentials: \[ dT = \frac{\partial T}{\partial x}dx + \frac{\partial T}{\partial y}dy + \frac{\partial T}{\partial z}dz \] First, we'll find the partial derivatives of \( T \) with respect to \( x \), \( y \), and \( z \). \[ \frac{\partial T}{\partial x} = y + z \] \[ \frac{\partial T}{\partial y} = x + z \] \[ \frac{\partial T}{\partial z} = x + y \] Now, we can substitute the coordinates of the first point (\( 2, -1, 3 \)) into the derivatives to get the rates of change at that point: \[ \frac{\partial T}{\partial x} = -1 + 3 = 2 \] \[ \frac{\partial T}{\partial y} = 2 + 3 = 5 \] \[ \frac{\partial T}{\partial z} = 2 - 1 = 1 \] Using the values of the partial derivatives at \( (2, -1, 3) \) and the differences in \( x \), \( y \), and \( z \) which are \( \Delta x = 1.98 - 2 = -0.02 \), \( \Delta y = -0.98 - (-1) = 0.02 \), and \( \Delta z = 3.02 - 3 = 0.02 \), the differential approximation of the change in \( T \) is: \[ dT \approx (2)(-0.02) + (5)(0.02) + (1)(0.02) \] \[ dT \approx -0.04 + 0.10 + 0.02 \] \[ dT \approx 0.08 \] So, according to the differential approximation, the temperature difference between the points \( (2, -1, 3) \) and \( (1.98, -0.98, 3.02) \) is approximately \( 0.08 \) units.
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