Consider the following: $\frac{1 - (3.03)^2}{(5.99)^2} - \frac{1 - 3^2}{6^2}$ . Find $z = f(x, y)$ where $f(x, y) = \frac{1 - x^2}{y^2}$ . Use the total differential to approximate the quantity.

Question :

Consider the following: $\frac{1 - (3.03)^2}{(5.99)^2} - \frac{1 - 3^2}{6^2}$ . find $z = f(x, y)$ where $f(x, y) = \frac{1 - x^2}{y^2}$ . use the total differential to approximate the quantity.

$Consider the following: \frac{1 - (3.03)^2}{(5.99)^2} - \frac{1 - 3^2}{6^2}. f | Doubtlet.com$

Solution:

Neetesh Kumar | October 30, 2024

Calculus Homework Help

Differential Approximation
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Step-by-step solution:

To approximate $f(3.03, 5.99) - f(3, 6)$ using the total differential, we set up the following approximation:

$dz = f_x \, dx + f_y \, dy$

where:

$x = 3$ , $y = 6$
$dx = 3.03 - 3 = 0.03$
$dy = 5.99 - 6 = -0.01$

Step 1: Calculate the Partial Derivatives $f_x$ and $f_y$

The function given is: $f(x, y) = \frac{1 - x^2}{y^2}$

Calculate $f_x$ :

$f_x = \frac{\partial}{\partial x} \left(\frac{1 - x^2}{y^2}\right) = \frac{-2x}{y^2}$

Substitute $x = 3$ and $y = 6$ :

$f_x(3, 6) = \frac{-2 \cdot 3}{6^2} = \frac{-6}{36} = -\frac{1}{6}$
Calculate $f_y$ :

$f_y = \frac{\partial}{\partial y} \left(\frac{1 - x^2}{y^2}\right) = \frac{-2(1 - x^2)}{y^3}$

Substitute $x = 3$ and $y = 6$ :

$f_y(3, 6) = \frac{-2(1 - 9)}{6^3} = \frac{-2(-8)}{216} = \frac{16}{216} = \frac{2}{27}$

Step 2: Apply the Total Differential

Now we apply the total differential formula:

$dz = f_x \, dx + f_y \, dy$

Substitute $f_x(3, 6) = -\frac{1}{6}$ , $f_y(3, 6) = \frac{2}{27}$ , $dx = 0.03$ , and $dy = -0.01$ :

$dz = \left(-\frac{1}{6}\right)(0.03) + \left(\frac{2}{27}\right)(-0.01)$

Calculate each term:

$dz = -\frac{0.03}{6} - \frac{0.02}{27}$

Compute $-\frac{0.03}{6}$ :

$-\frac{0.03}{6} = -0.005$
Compute $-\frac{0.02}{27}$ :

$-\frac{0.02}{27} \approx -0.00074$

Adding these results:

$dz \approx -0.005 - 0.00074 = -0.00574$

Final Answer

(Rounded to Three Decimal Places)

$dz \approx -0.006$

Thus, the approximation for $f(3.03, 5.99) - f(3, 6)$ is: $\boxed{-0.006}$

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Consider the following: 1−(3.03)2(5.99)2−1−3262\frac{1 - (3.03)^2}{(5.99)^2} - \frac{1 - 3^2}{6^2}(5.99)21−(3.03)2​−621−32​. Find z=f(x,y)z = f(x, y)z=f(x,y) where f(x,y)=1−x2y2f(x, y) = \frac{1 - x^2}{y^2}f(x,y)=y21−x2​. Use the total differential to approximate the quantity.