Find the slope of the line perpendicular to $y=-\frac{4}{15}x+15$

grabrovi0u
2022-10-21
Answered

Find the slope of the line perpendicular to $y=-\frac{4}{15}x+15$

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Kenley Rasmussen

Answered 2022-10-22
Author has **13** answers

The slope of the given line is $m=-\frac{4}{15}$

The negative reciprocal: $m\prime =-(-\frac{15}{4})=\frac{15}{4}$

So, the slope of the perpendicular line is $\frac{15}{4}$

The negative reciprocal: $m\prime =-(-\frac{15}{4})=\frac{15}{4}$

So, the slope of the perpendicular line is $\frac{15}{4}$

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${y}_{n+1}={y}_{n}$

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What is the solution in general for ${x}_{0}$, ${y}_{0}$, ${z}_{0}$ arbitrary?

It is intended to be solved using Jordan Normal Form of the Linear Algebra knowledge. I have no idea how to start, can anyone give a hint?

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Consider a 3rd order linear homogeneous DE of the form

$Lu={u}^{\u2034}+{a}_{2}(x){u}^{\u2033}+{a}_{1}(x){u}^{\prime}+{a}_{0}(x)u=f(x)\text{}\text{}\text{}\text{}(1)$

and for which ${u}_{1}={e}^{-x}$ and ${u}_{2}={e}^{-2x}$ are solutions to the homogeneous form of (1).

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My attempt:

When I think of stability, I immediately think of eigenvalues (nodes etc). Hence I reduced (1) into a system of linear equations:

$\begin{array}{rl}\frac{du}{dt}& =y,\\ \frac{dy}{dt}& =z,\\ \frac{dz}{dt}& =-{a}_{2}z-{a}_{1}y-{a}_{0}u.\end{array}$

This gives a corresponding matrix

$A=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -{a}_{0}& -{a}_{1}& -{a}_{2}\end{array}\right).$

But after working with this, I feel as if I'm not on the right track. Any advice would be greatly appreciated.

$Lu={u}^{\u2034}+{a}_{2}(x){u}^{\u2033}+{a}_{1}(x){u}^{\prime}+{a}_{0}(x)u=f(x)\text{}\text{}\text{}\text{}(1)$

and for which ${u}_{1}={e}^{-x}$ and ${u}_{2}={e}^{-2x}$ are solutions to the homogeneous form of (1).

Let $f(x)=10{e}^{-2x}$. Give an example of a form of ${a}_{2}$,${a}_{1}$ and ${a}_{0}$ such that (1) has a stable equilibrium point and an example such that (1) has no stable equilibrium point.

My attempt:

When I think of stability, I immediately think of eigenvalues (nodes etc). Hence I reduced (1) into a system of linear equations:

$\begin{array}{rl}\frac{du}{dt}& =y,\\ \frac{dy}{dt}& =z,\\ \frac{dz}{dt}& =-{a}_{2}z-{a}_{1}y-{a}_{0}u.\end{array}$

This gives a corresponding matrix

$A=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -{a}_{0}& -{a}_{1}& -{a}_{2}\end{array}\right).$

But after working with this, I feel as if I'm not on the right track. Any advice would be greatly appreciated.