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2 years ago

Express (5x+3)/((2x-3)(x+2)) in partial fractions

Mathematics

1 answer:

kolezko [41]2 years ago

4 0

Answer:

(5x+3)/((2x-3)(x+2)) = 3/(2x-3)+1/(x+2)

Step-by-step explanation:

Let

(5x+3)/((2x-3)(x+2)) = A/(2x-3)+B/(x+2)

5x+3=A(x+2)+B(2x-3)

5x+3=x(A+2B)+(2A-3B)

comparing x:

5=A+2B ... (1)

comparing constant:

3=2A-3B ... (2)

By solving:

A=3, B=1

(5x+3)/((2x-3)(x+2)) = 3/(2x-3)+1/(x+2)

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If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, i.e.

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$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta$

$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta$

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