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3 years ago

Write parametric equations of the line with the equation 4x-6y=-12 HELP FAST

Mathematics

2 answers:

Dmitry [639]3 years ago

7 0

Answer:

x = 3t

y = 2 + 2t

Step-by-step explanation:

Use any two points on the line:

Using intercepts would be easy

4x - 6y = -12

(-3,0) and (0,2)

Make a vector connecting these two: < 0 - -3, 2 - 0>

Direction vector: < 3, 2 >

Let the parameter be 't'

x = 0 + 3t

x = 3t

y = 2 + 2t

zimovet [89]3 years ago

4 0

Answer: x = t, y = 2t/3 + 2

Step-by-step explanation:

You want to write both x and y in terms of t

so if you rearrange 4x - 6y = -12 you get

-6y = -4x - 12

6y = 4x + 12

y = 2x/3 + 2

set x = t

and y = 2t/3 + 2

and that's it!

another one you could do is

x = t/2

y = t/3 + 2

or even

x = 3t/2

y = t + 2

i like this one the most because it looks more elegant

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Question:

Find the point (,) on the curve $y = \sqrt x$ that is closest to the point (3,0).

[To do this, first find the distance function between (,) and (3,0) and minimize it.]

Answer:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

Step-by-step explanation:

$y = \sqrt x$ can be represented as: $(x,y)$

Substitute $\sqrt x$ for $y$

$(x,y) = (x,\sqrt x)$

So, next:

Calculate the distance between $(x,\sqrt x)$ and $(3,0)$

Distance is calculated as:

$d = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2}$

So:

$d = \sqrt{(x-3)^2 + (\sqrt x - 0)^2}$

$d = \sqrt{(x-3)^2 + (\sqrt x)^2}$

Evaluate all exponents

$d = \sqrt{x^2 - 6x +9 + x}$

Rewrite as:

$d = \sqrt{x^2 + x- 6x +9 }$

$d = \sqrt{x^2 - 5x +9 }$

Differentiate using chain rule:

Let

$u = x^2 - 5x +9$

$\frac{du}{dx} = 2x - 5$

So:

$d = \sqrt u$

$d = u^\frac{1}{2}$

$\frac{dd}{du} = \frac{1}{2}u^{-\frac{1}{2}}$

Chain Rule:

$d' = \frac{du}{dx} * \frac{dd}{du}$

$d' = (2x-5) * \frac{1}{2}u^{-\frac{1}{2}}$

$d' = (2x - 5) * \frac{1}{2u^{\frac{1}{2}}}$

$d' = \frac{2x - 5}{2\sqrt u}$

Substitute: $u = x^2 - 5x +9$

$d' = \frac{2x - 5}{2\sqrt{x^2 - 5x + 9}}$

Next, is to minimize (by equating d' to 0)

$\frac{2x - 5}{2\sqrt{x^2 - 5x + 9}} = 0$

Cross Multiply

$2x - 5 = 0$

Solve for x

$2x =5$

$x = \frac{5}{2}$

Substitute $x = \frac{5}{2}$ in $y = \sqrt x$

$y = \sqrt{\frac{5}{2}}$

Split

$y = \frac{\sqrt 5}{\sqrt 2}$

Rationalize

$y = \frac{\sqrt 5}{\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$y = \frac{\sqrt {10}}{\sqrt 4}$

$y = \frac{\sqrt {10}}{2}$

Hence:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

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fgiga [73]

So some quick tips for inequalities:
**What you do to one side you HAVE TO do it to the other 2**
**If you ever divide a negative number past the inequality signs, the signs FLIP!!**
(e.g. 2>-5x>25)
-2/5 <x < -5

I hope that this has helped!

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