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Alina [70]
3 years ago
6

First answer gets 20 points and brainliest question

Mathematics
1 answer:
lina2011 [118]3 years ago
4 0
What's the question lol
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Find the point (,) on the curve =8 that is closest to the point (3,0). [To do this, first find the distance function between (,)
ELEN [110]

Question:

Find the point (,) on the curve y = \sqrt x that is closest to the point (3,0).

[To do this, first find the distance function between (,) and (3,0) and minimize it.]

Answer:

(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})

Step-by-step explanation:

y = \sqrt x can be represented as: (x,y)

Substitute \sqrt x for y

(x,y) = (x,\sqrt x)

So, next:

Calculate the distance between (x,\sqrt x) and (3,0)

Distance is calculated as:

d = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2}

So:

d = \sqrt{(x-3)^2 + (\sqrt x - 0)^2}

d = \sqrt{(x-3)^2 + (\sqrt x)^2}

Evaluate all exponents

d = \sqrt{x^2 - 6x +9 + x}

Rewrite as:

d = \sqrt{x^2 + x- 6x +9 }

d = \sqrt{x^2 - 5x +9 }

Differentiate using chain rule:

Let

u = x^2 - 5x +9

\frac{du}{dx} = 2x - 5

So:

d = \sqrt u

d = u^\frac{1}{2}

\frac{dd}{du} = \frac{1}{2}u^{-\frac{1}{2}}

Chain Rule:

d' = \frac{du}{dx} * \frac{dd}{du}

d' = (2x-5) * \frac{1}{2}u^{-\frac{1}{2}}

d' = (2x - 5) * \frac{1}{2u^{\frac{1}{2}}}

d' = \frac{2x - 5}{2\sqrt u}

Substitute: u = x^2 - 5x +9

d' = \frac{2x - 5}{2\sqrt{x^2 - 5x + 9}}

Next, is to minimize (by equating d' to 0)

\frac{2x - 5}{2\sqrt{x^2 - 5x + 9}} = 0

Cross Multiply

2x - 5 = 0

Solve for x

2x  =5

x = \frac{5}{2}

Substitute x = \frac{5}{2} in y = \sqrt x

y = \sqrt{\frac{5}{2}}

Split

y = \frac{\sqrt 5}{\sqrt 2}

Rationalize

y = \frac{\sqrt 5}{\sqrt 2} *  \frac{\sqrt 2}{\sqrt 2}

y = \frac{\sqrt {10}}{\sqrt 4}

y = \frac{\sqrt {10}}{2}

Hence:

(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})

3 0
3 years ago
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