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

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 $$\displaystyle{\left(\sqrt{{{2}}},{1}\right)}$$.

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Shannon Hodgkinson

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 $$\displaystyle\nabla{f}={2}\xi-{j}$$
$$\displaystyle\nabla{f}={2}\xi-{j}$$
$$\displaystyle={0}{i}-{j}$$
$$\displaystyle=-{j}$$

###### Not exactly what you’re looking for?
temnimam2
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 $$\displaystyle{y}={f{{\left({x}_{{0}}\right)}}}+{f}'{\left({x}_{{0}}\right)}{\left({x}-{x}_{{0}}\right)}$$
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}-{1}$$
$$\displaystyle{f{{\left(\sqrt{{{2}}}\right)}}}={\left(\sqrt{{{2}}}\right)}^{{2}}-{1}$$
$$\displaystyle={1}$$
$$\displaystyle{f}'{\left({x}\right)}={2}{x}$$
$$\displaystyle{f}'{\left(\sqrt{{{2}}}\right)}={2}\sqrt{{{2}}}$$
$$\displaystyle{y}={f{{\left(\sqrt{{{2}}}+{f}'{\left(\sqrt{{{2}}}\right)}{\left({x}-\sqrt{{{2}}}\right)}\right.}}}$$
$$\displaystyle={1}+{2}\sqrt{{{2}}}{\left({x}-\sqrt{{{2}}}\right)}$$
$$\displaystyle={2}\sqrt{{{2}}}{x}-{2}\sqrt{{{2}}}\cdot\sqrt{{{2}}}+{1}$$
$$\displaystyle={2}\sqrt{{{2}}}{x}-{3}$$