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Alecsey [184]
3 years ago
9

Y=1/2x-3Need it graphed only need the x and y intercept

Mathematics
2 answers:
MaRussiya [10]3 years ago
8 0
Here u gooo hope this helps

DIA [1.3K]3 years ago
8 0
Tha is how you graphed it

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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.

\dfrac{\mathrm d^2y}{\mathrm dx^2}

Compute the first derivative. By the chain rule,

\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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and so

\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

4 0
3 years ago
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