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bogdanovich [222]
3 years ago
11

The curve with equation y=ax^2+bx has a gradient of 3 at the point (2,-2). Find the values of a and b.

Mathematics
1 answer:
Airida [17]3 years ago
7 0

Answer:

a = 2, b = - 5

Step-by-step explanation:

Given

y = ax² + bx

\frac{dy}{dx} is the measure of the slope at x = a

Differentiate each term with respect to x using the power rule

\frac{d}{dx}(ax^{n} ) = nax^{n-1}

\frac{dy}{dx} = 2ax + b, hence

2ax + b = 3 at (2, - 2)

Substitute x = 2 into \frac{dy}{dx}

4a + b = 3 → (1) and substitute x = 2 into y

4a + 2b = - 2 → (2)

Subtract ( 1) from (2)

b = - 2 - 3 = - 5

Substitute b = - 5 into (1)

4a - 5 = 3 ( add 5 to both sides )

4a = 8 ( divide both sides by 4 )

a = 2

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Salt flows into the tank at a rate of

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and flows out at a rate of

(2 gal/min) (<em>s(t)</em>/300 lb/gal) = <em>s(t)</em>/150 lb/min

so that the net rate at which salt is exchanged through the tank is

d<em>s(t)</em>/d<em>t</em> = 12 - <em>s(t)</em>/150 … (lb/min)

Solve for <em>s(t)</em>. The DE is separable, so we have

d<em>s</em>/d<em>t</em> = 12 - <em>s</em>/150

150 d<em>s</em>/d<em>t</em> = 1800 - <em>s</em>

150/(1800 - <em>s</em>) d<em>s</em> = d<em>t</em>

Integrate both sides to get

-150 ln|1800 - <em>s</em>| = <em>t</em> + <em>C</em>

Solve for <em>s</em> :

ln|1800 - <em>s</em>| = -<em>t</em>/150 + <em>C</em>

1800 - <em>s</em> = exp(-<em>t</em>/150 + <em>C </em>)

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<em>s</em> = 1800 - <em>C</em> exp(-<em>t</em>/150)

Now given that <em>s(0)</em> = 50, we solve for <em>C</em> :

50 = 1800 - <em>C</em> exp(-0/150)

50 = 1800 - <em>C</em>

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To find the time it takes for the tank to hold 100 lb of salt, solve for <em>t</em> in

100 = 1800 - 1750 exp(-<em>t</em>/150)

1700 = 1750 exp(-<em>t</em>/150)

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ln(34/35) = -<em>t</em>/150

<em>t</em> = -150 ln(34/35) ≈ 4.348

So it would take approximately 4.348 minutes.

By the way, we didn't have to solve for <em>s(t)</em>, we could have instead stopped with

-150 ln|1800 - <em>s</em>| = <em>t</em> + <em>C</em>

Solve for <em>C</em> - this <em>C</em> <u>is not</u> the same as the one we found using the other method. <em>s(0)</em> = 50, so

-150 ln|1800 - 50| = 0 + <em>C</em>

<em>C</em> = -150 ln|1750|

==>   <em>t</em> = 150 ln(1750) - 150 ln|1800 - <em>s</em>|

Then <em>s(t)</em> = 100 lb when

<em>t</em> = 150 ln(1750) - 150 ln(1700)

<em>t</em> = 150 ln(1750/1700)

<em>t</em> = 150 ln(35/34)

<em>t</em> ≈ 4.348

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