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andreyandreev [35.5K]
2 years ago
8

Determine whether or not the vector field is conservative. If it is conservative, find a function f such that f = ∇f. (if the ve

ctor field is not conservative, enter dne. ) f(x, y, z) = i + sin(z)j + y cos(z)k.
SAT
1 answer:
Arisa [49]2 years ago
6 0

If F(x, y, z) = i + sin(z) j + y cos(z) k is conservative, then there exists a scalar function f(x, y, z) such that grad(f) = F, which means

∂f/∂x = 1

∂f/∂y = sin(z)

∂f/∂z = y cos(z)

Integrating each each of these equations gives

∫ ∂f/∂x dx = ∫ dx   ⇒   f(x, y, z) = x + α(y, z)

∫ ∂f/∂y dy = ∫ sin(z) dy   ⇒   f(x, y, z) = y sin(z) + β(x, z)

∫ ∂f/∂z dx = ∫ y cos(z) dz   ⇒   f(x, y, z) = y sin(z) + γ(x, y)

It follows that α(y, z) = y sin(z) and β(x, z) + γ(x, y) = x + C where C is a constant. So

f(x, y, z) = x + y sin(z) + C

and F is indeed conservative.

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Factor the expression into an equivalent form 12y^2-75.
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Hello oddworld7836!

\huge \boxed{\mathbb{QUESTION} \downarrow}

Factor the expression into an equivalent form 12y² - 75.

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

12 y ^ { 2 } - 75

By observing the expression, we can see that, 3 is the only common factor in both the terms of the expression. So, take the common factor 3 out.

12 y ^ { 2 } - 75 \\  = 3\left(4y^{2}-25\right)

Now, look at (4y² - 25). They don't have any common factors but they appear in the form of the algebraic identity ⇨ a² - b² = (a + b) (a - b). Here,

  • a² = 4, a = 2 (√a² = ✓4 = 2)
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So, the (4y² + 25) becomes...

(4 {y}^{2}  - 25) \\  = \left(2y-5\right)\left(2y+5\right)

Now, bring the 3 (common factor) & rewrite the complete expression.

12 y ^ { 2 } - 75 \\  =  \boxed{ \boxed{ \bf \: 3\left(2y-5\right)\left(2y+5\right) }}

We can't further simplify it. Also, remember that the simplified form of an expression is equivalent to the expression. So, 3 (2y - 5) (2y + 5) is equivalent to 12y² - 75.

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