Use the T-table below to determine 3 solutions to this equation: 4x - 2y = 8

1 answer:

5 0

To solve this problem, we need to pick numbers from the table then this number will be substituted into the equation. But first, we equate the numbers such as below:
4x - 2y = 8
Transpose -2y to the right side such as:
4x = 8 + 2y
4x - 8 = 2y
(4x -8)/2 = y
2x - 4 =y
From the table, let us choose x = 0
y = 2x -4
y = 2(0) -4
y = -4
From the table, let choose another value of x such as x=1
y = 2(1) -4
y = 2 - 4
y = -2

This will prove that the values in the table are correct.

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Find dy/dx and d2y/dx2. x = cos(2t), y = cos(t), 0 < t < π

pogonyaev

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}$

$y=\cos t\implies\dfrac{\mathrm dy}{\mathrm dt}=-\sin t$
$x=\cos2t\implies\dfrac{\mathrm dx}{\mathrm dt}=-2\sin2t=-4\sin t\cos t$
$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{-\sin t}{-4\sin t\cos t}=\dfrac14\sec t$

Let $z=\dfrac{\mathrm dy}{\mathrm dx}$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm dz}{\mathrm dx}=\dfrac{\mathrm dz}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac{\frac{\mathrm dz}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}$
$\implies\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac{\mathrm d}{\mathrm dt}\left[\frac14\sec t\right]}{-4\sin t\cos t}=\dfrac{\frac14\sec t\tan t}{-4\sin t\cos t}=-\dfrac1{16}\sec^3t$