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3 years ago

Use the change of variables s=x+3y, t=y to find the area of the ellipse x2+6xy+10y2≤1.

Mathematics

1 answer:

MArishka [77]3 years ago

5 0

$t=y\implies s=x+3t\implies x=s-3t$

Then

$x^2+6xy+10y^2=(s-3t)^2+6(s-3t)t+10t^2=s^2+t^2\le1$

which is a disk of radius 1, hence with area $\pi$ .

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PLZZ HELP

Digiron [165]

54 because there are 12 inches in a foot 12 divided by 3 is 4 and 6 divided by 3 is 2. Add the 2 and 4 and multiply that by 9

4 0

2 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6) Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

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2 years ago

SOLVE FOR X -5x + 9 = 24

boyakko [2]

Answer:

x = -3

Step-by-step explanation:

-5x + 9 = 24

move the nine to the other side

-5x = 15

divide

x = -3

7 0

3 years ago

(10-P) Help please, thanks.

Anuta_ua [19.1K]

The formula of linear equation is:
$y1 - y2 = m(x1 - x2)$
Where
x1 and x2: x coordinates(-1 and 3)
y1 and y2: y coordinates(-2 and 10)
m is the slope

We can then choose a point from the line to find the eqaution.

We need to first find the slope:
$\frac{y1 - y2}{x1 - x2}$
In this case:
$\frac{ - 2 - 10}{ - 1 - 3} \\ = \frac{ - 12}{ - 4} \\ = 3$

In this case, as the y2 is given as 1,put (-1,-2) , and the slope(3 )into the eqaution :

y-(-2) = 3(x-(-1))
y+2 = 3(x+1)

Therefore
$y + 2 = 3x + 1$
is the answer.

Hope it helps!

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3 years ago

Spark plugs should be charged every 15,000 miles. if a truck runs 5,000 miles a month,how many times a year should the spark plu

Furkat [3]

One month = 5000

Number of months before the spark plugs need to change :

15000 ÷ 5000 = 3

The sparks can last 3 months before it needs to be changed.

Number of times it need to change in a year = 12 ÷ 3 = 4

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Answer : It needs to be change 4 times a year.
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8 0

3 years ago