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AysviL [449]
3 years ago
8

Find the first partial derivatives of the function. f(x, y) = ax + by/ cx + dy

Mathematics
1 answer:
Marina86 [1]3 years ago
5 0
<span>The partial derivative of the given function with respect to x is a - by/cx2 + dy/dx In this derivative the terms in the with x is only considered other or treated as constant The partial derivative of the given function with respect to y is b/cx+ d2y/dx. In this derivative the terms in the with x is only considered other or treated as constant</span>
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What does a equal -5(6a+21)=-15
borishaifa [10]
Step 1 : Simplify both sides
-5(6a+21)=-15
(-5)(6a)+(-5)(21)=-15
-30a-105=-15
Step 2 : Add 105 to both sides
-30a-105+105=-15+105
-30a=90
Step 3 : Divide both sides by -30
-30a=90
—————
-30 -30

a = -3
7 0
3 years ago
37. Verify Green's theorem in the plane for f (3x2- 8y2) dx + (4y - 6xy) dy, where C is the boundary of the
Nastasia [14]

I'll only look at (37) here, since

• (38) was addressed in 24438105

• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

In both parts, we're considering the line integral

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy

and I assume <em>C</em> has a positive orientation in both cases

(a) It looks like the region has the curves <em>y</em> = <em>x</em> and <em>y</em> = <em>x</em> ² as its boundary***, so that the interior of <em>C</em> is the set <em>D</em> given by

D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}

• Compute the line integral directly by splitting up <em>C</em> into two component curves,

<em>C₁ </em>: <em>x</em> = <em>t</em> and <em>y</em> = <em>t</em> ² with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1

Then

\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}

*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.

• Compute the same integral using Green's theorem:

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}

(b) <em>C</em> is the boundary of the region

D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}

• Compute the line integral directly, splitting up <em>C</em> into 3 components,

<em>C₁</em> : <em>x</em> = <em>t</em> and <em>y</em> = 0 with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = <em>t</em> with 0 ≤ <em>t</em> ≤ 1

<em>C₃</em> : <em>x</em> = 0 and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1

Then

\displaystyle \int_C = \int_{C_1} + \int_{C_2} + \int_{C_3} \\\\ = \int_0^1 3t^2\,\mathrm dt + \int_0^1 (11t^2+4t-3)\,\mathrm dt + \int_0^1(4t-4)\,\mathrm dt \\\\ = \int_0^1 (14t^2+8t-7)\,\mathrm dt = \boxed{\frac53}

• Using Green's theorem:

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dx = \int_0^1\int_0^{1-x}10y\,\mathrm dy\,\mathrm dx = \boxed{\frac53}

4 0
3 years ago
Lemaitre (based on Hubble's observations) was the first to suggest the _____.
Molodets [167]
Universe exploded into existence 
6 0
3 years ago
Read 2 more answers
Right Triangle with legs of 15 and 20, what is the length of the Hypotenuse?
ioda
15^2 + 20^2
= 225 + 400
= 625
square root of 625 = 25

answer:<span>length of the Hypotenuse is</span> 25


5 0
3 years ago
How do you find the measure of ECF?
tigry1 [53]

Answer:

37°

Step-by-step explanation:

∠ECF= ∠ACG

∠ACG= 90° - ∠GAC= 90° - 53°= 37°

6 0
3 years ago
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