# S8.3.7.6 minimum vertical distance

• MHB
• karush
In summary, the question is asking for the minimum vertical distance between the parabolas $y=x^2+1$ and $y=x-x^2$. This can be found by finding the minimum of $|y_1-y_2|$ where $y_1=x^2+1$ and $y_2=x-x^2$. The minimum value is 15/16 when x=1/4, which can also be found by setting the derivative of the function 2x^2-x+1 equal to 0.
karush
Gold Member
MHB
S8.3.7.6. What is the minimum vertical distance between the parabolas
$$y = x^2+1 \textit{and } y = x- x^2$$

Ok I think what question is ... The vertical distance between vertex's

karush said:
S8.3.7.6. What is the minimum vertical distance between the parabolas
$$y = x^2+1 \textit{and } y = x- x^2$$

Ok I think what question is ... The vertical distance between vertex's
I interpret the question as the minimum of $|y_1 -y_2|$ where $y_1=x^2+1$ and $y_2=x-x^2$

... which would be a distance equal to 7/8

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$(x^2+1)'=2x$. Then 2x=0 so x=0
$(x-x^2)'=1-2x$ then 1-2x=0 so x=.5

(0)^2+1=1 And (.5)-(.5)^2=.25
Vertical distance is
$1-.25=.75$

karush said:
$(x^2+1)'=2x$. Then 2x=0 so x=0
$(x-x^2)'=1-2x$ then 1-2x=0 so x=.5

(0)^2+1=1 And (.5)-(.5)^2=.25
Vertical distance is
$1-.25=.75$

You calculated the vertical difference between the respective y-values of each vertex ... only problem is, the two vertices are not vertically aligned. The vertex of $y=x^2+1$ is at $(0,1)$ and the vertex of $y=x-x^2$ is $\left(\frac{1}{2},\frac{1}{4}\right)$

To get the idea, try the same question with these two functions ...

What is the minimum vertical distance between $y = x^2+1$ and $y = x-1$

... obviously, $y=x-1$ doesn't have a vertex.

Ok well this was from exercises in doing min/max problems where all the examples were solved by $f'(x)=0$

Yes the word distance was ackward as opposed to difference

This was an even problem # so no answer was given so you are probably correct

You are still ignoring skeeter's first response! No, this not asking for the distance between vertices, it is asking for the minimum of $$\displaystyle |y_1- y_2|= |x^2+ 1- x+ x^2|= |2x^2- x+ 1|$$. That (ignoring the absolute value) is a parabola, $$\displaystyle 2(x^2- x/2+ 1/16- 1/16)+ 1= 2(x- 1/4)^2+ 15/16$$. That's always positive so we can drop the absolute value. It has a minimum value of 15/16 when x= 1/4.

You could also use "f'(x)= 0" with f(x)= 2x^2-x+ 1. f'(x)= 4x- 1= 0. which gives x= 1/4 again.

## What is the minimum vertical distance?

The minimum vertical distance is the shortest distance between two points in a vertical direction. It is often measured from the ground level to the bottom of an object or from the top of an object to the ceiling.

## Why is minimum vertical distance important?

Minimum vertical distance is important for safety and compliance purposes. It ensures that there is enough space between objects to prevent accidents and allow for proper maintenance and operation.

## How is minimum vertical distance calculated?

The minimum vertical distance is calculated by measuring the vertical distance between two points using a measuring tool such as a ruler or tape measure. It can also be calculated using mathematical formulas, depending on the specific situation.

## What are some examples of minimum vertical distance requirements?

Some examples of minimum vertical distance requirements include the distance between a power line and a building, the distance between a ceiling and the top of a piece of machinery, and the distance between shelves in a warehouse.

## Are there any regulations or standards for minimum vertical distance?

Yes, there are regulations and standards set by organizations such as OSHA (Occupational Safety and Health Administration) and ANSI (American National Standards Institute) that specify minimum vertical distance requirements for different industries and situations.

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