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3 years ago

I don't fully understand this question if someone could lend me a hand that would be nice...

Mathematics

1 answer:

yulyashka [42]3 years ago

6 0

Answer:

y = 2327.49 ---> 25 hrs

starting plot points (0,640) and ending (25, 2327.49)

Step-by-step explanation:

b = 640

r = 5.3% = 0.053

t = 25

y = 640 ( 1 + 0.053)^25

y = 640 ( 1.053)^25

y = 2327.49 ---> 25 hrs

starting plot points (0,640) and ending (25, 2327.49)

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

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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

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

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

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

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

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

and so the second derivative is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta$

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