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# Find the point on the curve y=cosx closest to the point (0,0)?

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?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

3
Noah G Share
May 13, 2018

The point is $\left(0 , 1\right)$

#### Explanation:

By the distance formula, the distance between $\left(0 , 0\right)$ and the graph of $y = \cos x$ is

$d = \sqrt{{\left(x - 0\right)}^{2} + {\left(\cos x - 0\right)}^{2}}$

$d = \sqrt{{x}^{2} + {\cos}^{2} x}$

To find the minimum distance, we need to differentiate.

$d ' = \frac{2 x - 2 \cos x \sin x}{2 \sqrt{{x}^{2} + {\cos}^{2} x}}$

$d ' = \frac{x - \cos x \sin x}{\sqrt{{x}^{2} + {\cos}^{2} x}}$

We wish for this to be the smallest possible, thus we need $d ' = 0$.

$0 = \frac{x - \cos x \sin x}{\sqrt{{x}^{2} + {\cos}^{2} x}}$

$0 = x - \cos x \sin x$

Use a graphing application to solve and find that $x = 0$ is the only solution.

There derivative is negative when $x < 0$, and positive when $x > 0$, therefore, the minimum distance will occur when $x = 0$. The value of $\cos x$ at $x = 0$ is $1$, therefore the closest point is $\left(0 , 1\right)$. Let's examine graphically:

Hopefully this helps!

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