Respuesta :
sqrt(8)
The distance between some point of the curve and the origin can be calculated in such a way sqrt(x^2+(f(xW))^2)
We need to find the minimum of that function. Let's look at it's derivative:sqrt((x^2+(f(x))^2)′=x2−16/x2(√x4+16) The derivative equals zero at one point x=2/ It is the minimum as the sign changes from - to +. So the point we are tryinf to find is x=2 y=2. So the distanse is sqrt(8).
We need to find the minimum of that function. Let's look at it's derivative:sqrt((x^2+(f(x))^2)′=x2−16/x2(√x4+16) The derivative equals zero at one point x=2/ It is the minimum as the sign changes from - to +. So the point we are tryinf to find is x=2 y=2. So the distanse is sqrt(8).
The shortest distance [tex]x + y = 4[/tex] that is closest to the origin is [tex]\boxed{2\sqrt 2 {\text{ units}}}.[/tex]
Further explanation:
The formula for distance between the two points can be expressed as follows,
[tex]\boxed{{\text{Distance}} = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} }[/tex]
Given:
The line is [tex]x + y = 4.[/tex]
Explanation:
The coordinate of the origin is [tex]\left( {0,0} \right).[/tex]
The first point is [tex]\left( {x,y} \right)[/tex] and the second point is [tex]\left( {0,0} \right).[/tex]
The distance between the two points can be calculated as follows,
[tex]\begin{aligned}{\text{Distance}}&= \sqrt {{{\left( {x - 0} \right)}^2} + {{\left( {y - 0} \right)}^2}}\\&= \sqrt {{x^2} + {{\left( {4 - x} \right)}^2}}\\&= \sqrt {{x^2} + {x^2} - 8x + 16}\\&= \sqrt {2{x^2} - 8x + 16}\\\end{aligned}[/tex]
Differentiate the above equation with respect to [tex]x[/tex].
Substitute the first derivative equal to zero.
[tex]\begin{aligned}\frac{d}{{dx}}\left( {{\text{Distance}}} \right) &= 0\\\frac{{\left( {2x - 4} \right)}}{{\sqrt {2{x^2} - 8x + 16} }} &= 0\\2x - 4 &= 0\\2x &= 4\\x&= 2\\\end{aligned}[/tex]
Substitute [tex]x = 2[/tex] in equation [tex]x + y = 4[/tex] to obtain the value of [tex]y[/tex].
[tex]\begin{aligned}2 + y &= 4\\y&= 4 - 2\\y&= 2\\\end{aligned}[/tex]
The point is [tex]\left( {2,2} \right).[/tex]
The shortest distance can be obtained as follows,
[tex]\begin{aligned}{\text{Distance}} &= \sqrt {{2^2} + {2^2}}\\&= \sqrt {4 + 4}\\&= 2\sqrt 2\\\end{aligned}[/tex]
The shortest distance [tex]x + y = 4[/tex] that is closest to the origin is [tex]\boxed{2\sqrt 2 {\text{ units}}}.[/tex]
Learn more:
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Answer details:
Grade: High School
Subject: Mathematics
Chapter: Application of derivatives
Keywords: Derivative, shortest distance, curve, origin, attains, maximum, value of x, function, differentiate, minimum value, closest point, line, y+x = 4.
