Samacheer Kalvi Class 12 Maths Solutions Chapter 8 Differentials and Partial Derivatives Exercise 8.5

Chapter 8

Question 1. If \( w(x, y) = x^3 - 3xy + 2y^2 \), \( x, y \in \mathbb{R} \), find the linear approximation for w at (1, -1).
Answer: The function given is \( w(x, y) = x^3 - 3xy + 2y^2 \). We need to find its linear approximation at the point \( (x_0, y_0) = (1, -1) \). The linear approximation \( L(x, y) \) is given by the formula:
\( L(x, y) = w(x_0, y_0) + \left. \frac{\partial w}{\partial x} \right|_{(x_0, y_0)} (x - x_0) + \left. \frac{\partial w}{\partial y} \right|_{(x_0, y_0)} (y - y_0) \)

First, calculate \( w(x, y) \) at \( (1, -1) \):
\( w(1, -1) = (1)^3 - 3(1)(-1) + 2(-1)^2 \)
\( = 1 + 3 + 2 \)
\( = 6 \)

Next, find the partial derivatives of \( w(x, y) \) with respect to \( x \) and \( y \):
\( \frac{\partial w}{\partial x} = \frac{\partial}{\partial x} (x^3 - 3xy + 2y^2) = 3x^2 - 3y \)
\( \frac{\partial w}{\partial y} = \frac{\partial}{\partial y} (x^3 - 3xy + 2y^2) = -3x + 4y \)

Now, evaluate the partial derivatives at the point \( (1, -1) \):
\( \left. \frac{\partial w}{\partial x} \right|_{(1, -1)} = 3(1)^2 - 3(-1) = 3 + 3 = 6 \)
\( \left. \frac{\partial w}{\partial y} \right|_{(1, -1)} = -3(1) + 4(-1) = -3 - 4 = -7 \)

Substitute these values into the linear approximation formula:
\( L(x, y) = 6 + 6(x - 1) + (-7)(y + 1) \)
\( L(x, y) = 6 + 6x - 6 - 7y - 7 \)
\( L(x, y) = 6x - 7y - 7 \)
In simple words: We find a simple straight-line equation that closely matches the curvy function at a specific point. This is done by calculating the function's value and its slopes (partial derivatives) at that point, then putting them into a special formula.

🎯 Exam Tip: Remember that a linear approximation for a function of two variables creates a tangent plane at the given point, not just a tangent line.

Question 2. Let \( z(x, y) = x^2y + 3xy^4 \), \( x, y \in \mathbb{R} \). Find the linear approximation for z at (2, -1).
Answer: The function given is \( z(x, y) = x^2y + 3xy^4 \). We need to find its linear approximation at the point \( (x_0, y_0) = (2, -1) \). The linear approximation \( L(x, y) \) is given by:
\( L(x, y) = z(x_0, y_0) + \left. \frac{\partial z}{\partial x} \right|_{(x_0, y_0)} (x - x_0) + \left. \frac{\partial z}{\partial y} \right|_{(x_0, y_0)} (y - y_0) \)

First, calculate \( z(x, y) \) at \( (2, -1) \):
\( z(2, -1) = (2)^2(-1) + 3(2)(-1)^4 \)
\( = 4(-1) + 6(1) \)
\( = -4 + 6 \)
\( = 2 \)

Next, find the partial derivatives of \( z(x, y) \) with respect to \( x \) and \( y \):
\( \frac{\partial z}{\partial x} = \frac{\partial}{\partial x} (x^2y + 3xy^4) = 2xy + 3y^4 \)
\( \frac{\partial z}{\partial y} = \frac{\partial}{\partial y} (x^2y + 3xy^4) = x^2 + 12xy^3 \)

Now, evaluate the partial derivatives at the point \( (2, -1) \):
\( \left. \frac{\partial z}{\partial x} \right|_{(2, -1)} = 2(2)(-1) + 3(-1)^4 = -4 + 3(1) = -4 + 3 = -1 \)
\( \left. \frac{\partial z}{\partial y} \right|_{(2, -1)} = (2)^2 + 12(2)(-1)^3 = 4 + 24(-1) = 4 - 24 = -20 \)

Substitute these values into the linear approximation formula:
\( L(x, y) = 2 + (-1)(x - 2) + (-20)(y + 1) \)
\( L(x, y) = 2 - (x - 2) - 20(y + 1) \)
\( L(x, y) = 2 - x + 2 - 20y - 20 \)
\( L(x, y) = -x - 20y - 16 \)
\( L(x, y) = -(x + 20y + 16) \)
In simple words: We are finding a flat surface (a plane) that just touches the 3D graph of the function at a certain point. This plane is the best straight-line estimate of the function near that point.

🎯 Exam Tip: Be careful with signs, especially when substituting negative values for x and y into the partial derivatives and the main formula.

Question 3. If \( v(x, y) = x^2 - xy + \frac{1}{4}y^2 + 7 \), \( x, y \in \mathbb{R} \), find the differential dv.
Answer: The function given is \( v(x, y) = x^2 - xy + \frac{1}{4}y^2 + 7 \). To find the total differential \( dv \), we first need to calculate the partial derivatives of \( v \) with respect to \( x \) and \( y \). The total differential formula for a two-variable function \( v(x, y) \) is:
\( dv = \frac{\partial v}{\partial x} dx + \frac{\partial v}{\partial y} dy \)

First, find the partial derivative with respect to \( x \):
\( \frac{\partial v}{\partial x} = \frac{\partial}{\partial x} (x^2 - xy + \frac{1}{4}y^2 + 7) \)
\( = 2x - y \)

Next, find the partial derivative with respect to \( y \):
\( \frac{\partial v}{\partial y} = \frac{\partial}{\partial y} (x^2 - xy + \frac{1}{4}y^2 + 7) \)
\( = -x + \frac{1}{2}y \)

Finally, substitute these partial derivatives into the formula for \( dv \):
\( dv = (2x - y) dx + (-x + \frac{1}{2}y) dy \)
In simple words: The differential \( dv \) tells us how much the function \( v \) changes when \( x \) and \( y \) change by very small amounts, \( dx \) and \( dy \). It uses the slopes in the \( x \) and \( y \) directions.

🎯 Exam Tip: Remember to apply differentiation rules correctly to each term when finding partial derivatives, treating other variables as constants.

Question 4. Let \( W(x, y, z) = x^2 - xy + 3\sin z \), \( x, y, z \in \mathbb{R} \). Find the linear approximation at (2, -1, 0).
Answer: The function given is \( W(x, y, z) = x^2 - xy + 3\sin z \). We need to find its linear approximation at the point \( (x_0, y_0, z_0) = (2, -1, 0) \). The linear approximation \( L(x, y, z) \) for a three-variable function is given by:
\( L(x, y, z) = W(x_0, y_0, z_0) + \left. \frac{\partial W}{\partial x} \right|_{(x_0, y_0, z_0)} (x - x_0) + \left. \frac{\partial W}{\partial y} \right|_{(x_0, y_0, z_0)} (y - y_0) + \left. \frac{\partial W}{\partial z} \right|_{(x_0, y_0, z_0)} (z - z_0) \)

First, calculate \( W(x, y, z) \) at \( (2, -1, 0) \):
\( W(2, -1, 0) = (2)^2 - (2)(-1) + 3\sin(0) \)
\( = 4 + 2 + 3(0) \)
\( = 6 \)
(Since \( \sin(0) = 0 \)).

Next, find the partial derivatives of \( W(x, y, z) \) with respect to \( x, y, \) and \( z \):
\( \frac{\partial W}{\partial x} = \frac{\partial}{\partial x} (x^2 - xy + 3\sin z) = 2x - y \)
\( \frac{\partial W}{\partial y} = \frac{\partial}{\partial y} (x^2 - xy + 3\sin z) = -x \)
\( \frac{\partial W}{\partial z} = \frac{\partial}{\partial z} (x^2 - xy + 3\sin z) = 3\cos z \)

Now, evaluate the partial derivatives at the point \( (2, -1, 0) \):
\( \left. \frac{\partial W}{\partial x} \right|_{(2, -1, 0)} = 2(2) - (-1) = 4 + 1 = 5 \)
\( \left. \frac{\partial W}{\partial y} \right|_{(2, -1, 0)} = -(2) = -2 \)
\( \left. \frac{\partial W}{\partial z} \right|_{(2, -1, 0)} = 3\cos(0) = 3(1) = 3 \)
(Since \( \cos(0) = 1 \)).

Substitute these values into the linear approximation formula:
\( L(x, y, z) = 6 + 5(x - 2) + (-2)(y + 1) + 3(z - 0) \)
\( L(x, y, z) = 6 + 5x - 10 - 2y - 2 + 3z \)
\( L(x, y, z) = 5x - 2y + 3z - 6 \)
In simple words: This is like finding a flat plane that touches the 3D surface of the function at one exact point. This plane helps us estimate the function's value for points very close to that contact point.

🎯 Exam Tip: For trigonometric functions, remember the values of common angles like 0, \( \frac{\pi}{2} \), \( \pi \) for sine and cosine when evaluating partial derivatives.

Question 5. Let \( V(x, y, z) = xy + yz + zx \), \( x, y, z \in \mathbb{R} \). Find the differential dV.
Answer: The function given is \( V(x, y, z) = xy + yz + zx \). To find the total differential \( dV \), we need to calculate the partial derivatives of \( V \) with respect to \( x, y, \) and \( z \). The total differential formula for a three-variable function \( V(x, y, z) \) is:
\( dV = \frac{\partial V}{\partial x} dx + \frac{\partial V}{\partial y} dy + \frac{\partial V}{\partial z} dz \)

First, find the partial derivative with respect to \( x \):
\( \frac{\partial V}{\partial x} = \frac{\partial}{\partial x} (xy + yz + zx) = y + z \)

Next, find the partial derivative with respect to \( y \):
\( \frac{\partial V}{\partial y} = \frac{\partial}{\partial y} (xy + yz + zx) = x + z \)

Then, find the partial derivative with respect to \( z \):
\( \frac{\partial V}{\partial z} = \frac{\partial}{\partial z} (xy + yz + zx) = y + x \)

Finally, substitute these partial derivatives into the formula for \( dV \):
\( dV = (y + z) dx + (x + z) dy + (y + x) dz \)
In simple words: The total differential \( dV \) shows how a function that depends on three things (\( x, y, z \)) changes when each of those things changes just a little bit. It adds up the small changes from each direction.

🎯 Exam Tip: When finding partial derivatives, remember to treat other variables as constants. For example, when finding \( \frac{\partial V}{\partial x} \), \( y \) and \( z \) are treated as fixed numbers.