Let \({\rm{f}}\left( {{\rm{x}},{\rm{y}}} \right) = \frac{{{\rm{a}}{{\rm{x}}^2} + {\rm{b}}{{\rm{y}}^2}}}{{{\rm{xy}}}}\), where a and b are constants. If \(\frac{{\partial {\rm{f}}}}{{\partial {\rm{x}}}} = \frac{{\partial {\rm{f}}}}{{\partial {\rm{y}}}}{\rm{\;}}\) at x = 1 and y = 2, then the relation between a and b is
a = 4b
To find the relation between constants \(a\) and \(b\), we need to calculate the partial derivatives of the given function \(f(x,y)\) with respect to \(x\) and \(y\), then evaluate them at the specified point \((x,y) = (1,2)\), and finally equate them as per the problem statement.
The given function is: $$f(x,y) = \frac{ax^2 + by^2}{xy}$$ We can rewrite the function for easier differentiation:
$$f(x,y) = \frac{ax^2}{xy} + \frac{by^2}{xy} = \frac{ax}{y} + \frac{by}{x}$$ $$f(x,y) = ax y^{-1} + by x^{-1}$$Partial Derivatives Calculation
Let's calculate the partial derivatives:
Derivatives with Respect to x
To find the partial derivative of \(f\) with respect to \(x\) (\(\frac{{\partial {\rm{f}}}}{{\partial {\rm{x}}}}\)), we treat \(y\) as a constant:
- Differentiate \(ax y^{-1}\) with respect to \(x\): \(a y^{-1} \cdot \frac{d}{dx}(x) = a y^{-1} \cdot 1 = \frac{a}{y}\)
- Differentiate \(by x^{-1}\) with respect to \(x\): \(by \cdot \frac{d}{dx}(x^{-1}) = by \cdot (-1)x^{-2} = -\frac{by}{x^2}\)
So, the partial derivative with respect to \(x\) is:
$$\frac{{\partial {\rm{f}}}}{{\partial {\rm{x}}}} = \frac{a}{y} - \frac{by}{x^2}$$Derivatives with Respect to y
To find the partial derivative of \(f\) with respect to \(y\) (\(\frac{{\partial {\rm{f}}}}{{\partial {\rm{y}}}}\)), we treat \(x\) as a constant:
- Differentiate \(ax y^{-1}\) with respect to \(y\): \(ax \cdot \frac{d}{dy}(y^{-1}) = ax \cdot (-1)y^{-2} = -\frac{ax}{y^2}\)
- Differentiate \(by x^{-1}\) with respect to \(y\): \(b x^{-1} \cdot \frac{d}{dy}(y) = b x^{-1} \cdot 1 = \frac{b}{x}\)
So, the partial derivative with respect to \(y\) is:
$$\frac{{\partial {\rm{f}}}}{{\partial {\rm{y}}}} = -\frac{ax}{y^2} + \frac{b}{x}$$Evaluating Derivatives at Given Point
The problem states that \(\frac{{\partial {\rm{f}}}}{{\partial {\rm{x}}}} = \frac{{\partial {\rm{f}}}}{{\partial {\rm{y}}}}\) at \(x = 1\) and \(y = 2\). Let's substitute these values into our partial derivative expressions.
Evaluation of \(\frac{{\partial {\rm{f}}}}{{\partial {\rm{x}}}}\)
At \(x = 1\) and \(y = 2\):
$$\left. \frac{{\partial {\rm{f}}}}{{\partial {\rm{x}}}} \right|_{(1,2)} = \frac{a}{2} - \frac{b(2)}{(1)^2}$$ $$\left. \frac{{\partial {\rm{f}}}}{{\partial {\rm{x}}}} \right|_{(1,2)} = \frac{a}{2} - 2b$$Evaluation of \(\frac{{\partial {\rm{f}}}}{{\partial {\rm{y}}}}\)
At \(x = 1\) and \(y = 2\):
$$\left. \frac{{\partial {\rm{f}}}}{{\partial {\rm{y}}}} \right|_{(1,2)} = -\frac{a(1)}{(2)^2} + \frac{b}{1}$$ $$\left. \frac{{\partial {\rm{f}}}}{{\partial {\rm{y}}}} \right|_{(1,2)} = -\frac{a}{4} + b$$Relation Between a and b
Now, we set the evaluated partial derivatives equal to each other:
$$\frac{a}{2} - 2b = -\frac{a}{4} + b$$To eliminate the fractions, multiply the entire equation by the least common multiple of the denominators (2 and 4), which is 4:
$$4 \left( \frac{a}{2} - 2b \right) = 4 \left( -\frac{a}{4} + b \right)$$ $$2a - 8b = -a + 4b$$Now, we rearrange the terms to group \(a\) terms on one side and \(b\) terms on the other side:
$$2a + a = 4b + 8b$$ $$3a = 12b$$Finally, divide both sides by 3 to find the relation between \(a\) and \(b\):
$$\frac{3a}{3} = \frac{12b}{3}$$ $$a = 4b$$This is the required relation between \(a\) and \(b\).
The final answer is a = 4b.
Important Questions from Differentiability
What is the value of f'(x) at x = 4 from the following table of values?
x 1 2 3 4 f(x) 20 22 27 35 The set of all points, where the function \({\rm{f}}\left( {\rm{x}} \right) = \sqrt {1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}} \) is differentiable, is
Let f be a differentiable function defined for all x ∈ R such that f(x3) = x5 for all x ∈ R, x ≠ 0. Then the value of \(\dfrac{df}{dx} (8)\) is:
If \(f(x)=\displaystyle\sum_{n-0}^{2k}\left(a_n|x|^n+b_n\ \sin^2x\right)\), where \(a_i^{'}\)s and \(b_i^{'}\)s (0 ≤ i ≤ k) are real constants, then f(x) is:
The set of all point where the function f(x) = 2x|x| is differentiable, is:
