garnentas3m

2022-01-04

Draw the gradient $\mathrm{\nabla }f$ and level curves for $f\left(x,y\right)={x}^{2}-y$ at a few common locations. For the point where the level curve $f\left(x,y\right)=1$ meets the tangent line, write an equation $\left(\sqrt{2},1\right)$.

Shannon Hodgkinson

Step 1
If $y=f\left(x\right)$ , 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 $f\left(x,y,z\right)$ contains more than one inputs.
Step 2
The gradient of a multivariable function $f\left(x,y\right)$ is defined as $\mathrm{\nabla }f-\frac{\partial f}{\partial x}i+\frac{\partial f}{\partial y}j$. To find gradient of $f\left(x,y\right)={x}^{2}-y$, substitute it in $\mathrm{\nabla }f-\frac{\partial f}{\partial x}i+\frac{\partial f}{\partial y}j$. According to standard results in partial differentiation $\frac{\partial {x}^{n}}{\partial x}=n{x}^{n-1}$ and $\frac{\partial {y}^{n}}{\partial x}=n{y}^{n-1}$, where n is an integer.
$\mathrm{\nabla }f=\frac{\partial f}{\partial x}i+\frac{\partial f}{\partial y}j$
$=i\frac{\partial }{\partial x}\left({x}^{2}-y\right)+j\frac{\partial }{\partial y}\left({x}^{2}-y\right)$
$=2\xi -j$
To find the gradient at a particular point $\left({x}_{0},{y}_{0}\right)$, substitute it in $\mathrm{\nabla }f=2\xi -j$ 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 $y=0$ in $\mathrm{\nabla }f=2\xi -j$
$\mathrm{\nabla }f=2\xi -j$
$=0i-j$
$=-j$

temnimam2

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

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