Sketch the level curves for f(x,y) = x^2-y together with

garnentas3m 2022-01-04 Answered
Sketch the level curves for \(\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{2}}-{y}\) together with the gradient \(\displaystyle\nabla{f}\) at a few typical points. Write an equation for the tangent line of the level curve \(\displaystyle{f{{\left({x},{y}\right)}}}={1}\) at the point \(\displaystyle{\left(\sqrt{{{2}}},{1}\right)}\).

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Expert Answer

Shannon Hodgkinson
Answered 2022-01-05 Author has 2714 answers

Step 1
If \(\displaystyle{y}={f{{\left({x}\right)}}}\) , then the independent variable x is called the input of the function and the dependent variable y is called the output of the function. The derivative of the function is the measure of how fast the function is changing with respect to x. A multi variable function \(\displaystyle{f{{\left({x},{y},{z}\right)}}}\) contains more than one inputs.
Step 2
The gradient of a multivariable function \(\displaystyle{f{{\left({x},{y}\right)}}}\) is defined as \(\displaystyle\nabla{f}-{\frac{{\partial{f}}}{{\partial{x}}}}{i}+{\frac{{\partial{f}}}{{\partial{y}}}}{j}\). To find gradient of \(\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{2}}-{y}\), substitute it in \(\nabla f-\frac{\partial f}{\partial x}i+\frac{\partial f}{\partial y}j\). According to standard results in partial differentiation \(\displaystyle{\frac{{\partial{x}^{{n}}}}{{\partial{x}}}}={n}{x}^{{{n}-{1}}}\) and \(\displaystyle{\frac{{\partial{y}^{{n}}}}{{\partial{x}}}}={n}{y}^{{{n}-{1}}}\), where n is an integer.
\(\displaystyle\nabla{f}={\frac{{\partial{f}}}{{\partial{x}}}}{i}+{\frac{{\partial{f}}}{{\partial{y}}}}{j}\)
\(\displaystyle={i}{\frac{{\partial}}{{\partial{x}}}}{\left({x}^{{2}}-{y}\right)}+{j}{\frac{{\partial}}{{\partial{y}}}}{\left({x}^{{2}}-{y}\right)}\)
\(\displaystyle={2}\xi-{j}\)
To find the gradient at a particular point \(\displaystyle{\left({x}_{{0}},{y}_{{0}}\right)}\), substitute it in \(\displaystyle\nabla{f}={2}\xi-{j}\) and find the corresponding vectors. Thus the value of gradient corresponding to the points that lies on the y axis can be calculated by substituting \(\displaystyle{y}={0}\) in \(\displaystyle\nabla{f}={2}\xi-{j}\)
\(\displaystyle\nabla{f}={2}\xi-{j}\)
\(\displaystyle={0}{i}-{j}\)
\(\displaystyle=-{j}\)

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temnimam2
Answered 2022-01-06 Author has 3589 answers
Second part:
The equation of the tangent line to the function \(\displaystyle{y}={f{{\left({x}\right)}}}\) at \(\displaystyle{x}={x}_{{0}}\) is given by \(\displaystyle{y}={f{{\left({x}_{{0}}\right)}}}+{f}'{\left({x}_{{0}}\right)}{\left({x}-{x}_{{0}}\right)}\). In this problem the \(\displaystyle{f{{\left({x},{y}\right)}}}={1}\), since \(\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{2}}-{y}\) the equation of curve in the x-y plane becomes \(\displaystyle{x}^{{2}}-{y}={1}\) or \(\displaystyle{y}={x}^{{2}}-{1}\).
To find the equation of the tangent line to the curve \(\displaystyle{y}={x}^{{2}}-{1}\) at \(\displaystyle{\left(\sqrt{{{2}}},{1}\right)}\), find the value of \(\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}-{1}\) and its derivative at \(\displaystyle{x}=\sqrt{{{2}}}\) and then substitute obtained values in \(\displaystyle{y}={f{{\left({x}_{{0}}\right)}}}+{f}'{\left({x}_{{0}}\right)}{\left({x}-{x}_{{0}}\right)}\)
\(\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}-{1}\)
\(\displaystyle{f{{\left(\sqrt{{{2}}}\right)}}}={\left(\sqrt{{{2}}}\right)}^{{2}}-{1}\)
\(\displaystyle={1}\)
\(\displaystyle{f}'{\left({x}\right)}={2}{x}\)
\(\displaystyle{f}'{\left(\sqrt{{{2}}}\right)}={2}\sqrt{{{2}}}\)
\(\displaystyle{y}={f{{\left(\sqrt{{{2}}}+{f}'{\left(\sqrt{{{2}}}\right)}{\left({x}-\sqrt{{{2}}}\right)}\right.}}}\)
\(\displaystyle={1}+{2}\sqrt{{{2}}}{\left({x}-\sqrt{{{2}}}\right)}\)
\(\displaystyle={2}\sqrt{{{2}}}{x}-{2}\sqrt{{{2}}}\cdot\sqrt{{{2}}}+{1}\)
\(\displaystyle={2}\sqrt{{{2}}}{x}-{3}\)
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