How do you find a unit vector (a) parallel to and (b) normal to the graph of f(x) at the indicated points given function: $f (x) = \sqrt{25 + x^{2}}$ $f(x) = sqrt(25+x^2)$ and point: (3,4)?

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How do you find a unit vector (a) parallel to and (b) normal to the graph of f(x) at the indicated points given function: $f (x) = \sqrt{25 + x^{2}}$ $f(x) = sqrt(25+x^2)$ and point: (3,4)?

1 Answer

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Answer 1 · 2018-07-26T18:45:05

Unit vector parallel to the graph at (3,4) is is
$n^{\to} = \frac{4}{5} i + \frac{3}{5} j$ $n^->=4/5i +3/5j^$
Unit vector normal to the graph at (3,4) is is
$n^{\to} = \frac{3}{5} i - \frac{4}{5} j$ $n^->=3/5i-4/5j$

Explanation:

$f (x) = \sqrt{25 + x^{2}}$ $f(x)=sqrt(25+x^2)$
slope of the tangent indicates a vector parallel to the graph at a point
Differentiating
Let
$y = f (x)$ $y=f(x)$
$y = \sqrt{25 + x^{2}}$ $y=sqrt(25+x^2)$
Squaring both sides
$y^{2} = 25 + x^{2}$ $y^2=25+x^2$
$y^{2} - x^{2} = 5^{2}$ $y^2-x^2=5^2$
Now, Applyig chain rule and differentiating
$2 y \frac{d y}{d x} - 2 x = 0$ $2ydy/dx-2x=0$
$\frac{d y}{d x} = \frac{x}{y}$ $dy/dx=x/y$
At
$(x, y) \equiv (3, 4);$ $(x,y)-=(3,4);$
$\frac{d y}{d x} = \frac{3}{4}$ $dy/dx=3/4$
Slope of a parallel line is
$\sqrt{3^{2} + 4^{2}} = 5$ $sqrt(3^2+4^2)=5$
Unit vector parallel to the graph at (3,4) is is
$n^{\to} = \frac{4}{5} i + \frac{3}{5} j$ $n^->=4/5i +3/5j^$
Normal is perpendicular to the parallel.
Thus, slope of the normal is
$m_{1} m_{2} = - 1$ $m_1m_2=-1$
Slope of anormal line is
$- \frac{4}{3}$ $-4/3$
Unit vector normal to the graph at (3,4) is is
$n^{\to} = \frac{3}{5} i - \frac{4}{5} j$ $n^->=3/5i-4/5j$

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How do you find a unit vector (a) parallel to and (b) normal to the graph of f(x) at the indicated points given function: $f (x) = \sqrt{25 + x^{2}}$ $f(x) = sqrt(25+x^2)$ and point: (3,4)?

1 Answer

Explanation:

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How do you find a unit vector (a) parallel to and (b) normal to the graph of f(x) at the indicated points given function: f(x)=√25+x2f(x) = sqrt(25+x^2) and point: (3,4)?

1 Answer

Explanation:

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How do you find a unit vector (a) parallel to and (b) normal to the graph of f(x) at the indicated points given function: $f (x) = \sqrt{25 + x^{2}}$ $f(x) = sqrt(25+x^2)$ and point: (3,4)?