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Create the local linearization of xy=8 at x=2.5.

User Cem U
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1 Answer

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Answer:

To find the local linearization of the equation \(xy = 8\) at \(x = 2.5\), we'll use the concept of linear approximation, which involves finding the equation of the tangent line to the curve at the given point. The equation of a line in point-slope form is given by:

\[y - y_1 = m(x - x_1)\]

Where:

- \(y\) and \(x\) are the variables in question.

- \((x_1, y_1)\) is the point we want to approximate around.

- \(m\) is the slope of the tangent line.

Let's first find the slope, \(m\), at the point \((2.5, y_1)\). To do this, we'll need to find both \(\frac{dy}{dx}\) and evaluate the equation at \(x = 2.5\).

Given the equation \(xy = 8\), we can implicitly differentiate it with respect to \(x\):

\[\frac{d(xy)}{dx} = \frac{d(8)}{dx}\]

Using the product rule for differentiation, we get:

\[x\frac{dy}{dx} + y\frac{dx}{dx} = 0\]

Since \(\frac{dx}{dx} = 1\), we can simplify to:

\[x\frac{dy}{dx} + y = 0\]

Now, we need to evaluate this at \(x = 2.5\). Plugging in \(x = 2.5\) gives us:

\[2.5\frac{dy}{dx} + y_1 = 0\]

Now, we need to solve for \(\frac{dy}{dx}\), the slope of the tangent line at \(x = 2.5\):

\[2.5\frac{dy}{dx} = -y_1\]

\[\frac{dy}{dx} = -\frac{y_1}{2.5}\]

Next, we'll evaluate the original equation \(xy = 8\) at \(x = 2.5\) to find \(y_1\):

\[2.5y_1 = 8\]

Now, solve for \(y_1\):

\[y_1 = \frac{8}{2.5} = 3.2\]

So, the point \((2.5, y_1)\) is \((2.5, 3.2)\), and we've already found the slope \(\frac{dy}{dx} = -\frac{3.2}{2.5} = -1.28\).

Now, we can write the equation of the tangent line using point-slope form:

\[y - 3.2 = -1.28(x - 2.5)\]

Simplify and rearrange:

\[y - 3.2 = -1.28x + 3.2\]

Now, add \(3.2\) to both sides:

\[y = -1.28x + 6.4\]

So, the local linearization of the curve \(xy = 8\) at \(x = 2.5\) is \(y = -1.28x + 6.4\).

Explanation:

User Nithin Girish
by
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