S8.5.1.64 values of m for region

[karush]

$\tiny{s8.5.1.64\p364}$

For what values of m do the line $y=mx$ and the curve $y=\dfrac{x}{x^2+1}$ enclose a region.
Find the area of the regionok I could only estimate this by observation but it looks $m\ne 1$
not sure how you solve by calculation

 

[skeeter]

both functions are odd (symmetric to the origin) and pass through the origin. the curve is also asymptotic to $y=0$.

$\dfrac{d}{dx} \bigg[\dfrac{x}{x^2+1} \bigg] = \dfrac{1-x^2}{(x^2+1)^2}$

at the origin, the slope of the curve equals 1, therefore $y=x$ would be tangent to the curve.

so, to intersect the curve to form closed symmetrical regions in quadrants I and III, the slope of the line must be $0 < m < 1$

 

[karush]

ok that helped a lot

 

[HOI]

$\tiny{s8.5.1.64\p364}$

For what values of m do the line $y=mx$ and the curve $y=\dfrac{x}{x^2+1}$ enclose a region.
Find the area of the regionok I could only estimate this by observation but it looks $m\ne 1$
not sure how you solve by calculation

Since the problem says "enclose a region" I would just see where they intersect:
$mx= \frac{x}{x^2+ 1}$
An obvious solution is x= 0. If x is not 0 we can divide by x to get $m= \frac{1}{x^2+ 1}$. Multiply both sides by $x^2+ 1$ to get $m(x^2+ 1)= 1$. If m= 0 that is impossible and as long as m is not 0, divide both sides by m to get $x^2+ 1= \frac{1}{m}$. Subtract $1$: $x^2= \frac{1}{m}- 1$. If m> 1 there is no such x. If m< 1 $x= \pm\sqrt{\frac{1}{m}- 1}= \pm\sqrt{\frac{1- m}{m}}$ so the lines cross in three points and enclose two regions. The answer is "all m less than 1" not "all m except 1".

 

[skeeter]

$\tiny{s8.5.1.64\p364}$

For what values of m do the line $y=mx$ and the curve $y=\dfrac{x}{x^2+1}$ enclose a region.

Find the area of the region

for $x > 0$ and $0 < m < 1$ ...

$mx = \dfrac{x}{x^2+1} \implies x^2+1 = \dfrac{1}{m} \implies x = \sqrt{\dfrac{1}{m} - 1}$

Sum of the areas of both regions in quads I and III ...

\(\displaystyle A = 2 \int_0^{\sqrt{\frac{1}{m}-1}} \dfrac{x}{x^2+1} - mx \, dx = m - (1+\ln{m})\)

 

[karush]

both functions are odd (symmetric to the origin) and pass through the origin. the curve is also asymptotic to $y=0$.

$\dfrac{d}{dx} \bigg[\dfrac{x}{x^2+1} \bigg] = \dfrac{1-x^2}{(x^2+1)^2}$

at the origin, the slope of the curve equals 1, therefore $y=x$ would be tangent to the curve.

so, to intersect the curve to form closed symmetrical regions in quadrants I and III, the slope of the line must be $0 < m < 1$

View attachment 10706

well pretty valuable to know that...