Mathematics, 16.09.2020 20:01 twiddleturd

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Answer from: corbinfisher

Let's consider a normal line that is perpendicular to the curve in the first quadrant at the abcissa

. Being locally perpendicular to the curve, its slope is perpendicular to the curve's local slope, which is its derivative at

. The derivative of x^2

hence the curve's local slope at that point is

. As a result, the normal line's slope is $-\frac{1}{2a}$ . The normal line's curve's equation is thus of the form $-\frac{1}{2a}x+b$ . What is more, its value at

hence $a^2=-\frac{1}{2a}a+b$ thus $b=x_0^2+\frac{1}2$ . Hence the equation of the normal line is $\boxed{y=-\frac{1}{2a}x+a^2+\frac{1}2}$

Let's proceed to finding where it intercept the curve in the second quadrant. We have to solve $x^2=-\frac{1}{2a}x+a^2+\frac{1}2$ . This equation's discriminant is $\Delta=\frac{1}{4a^2}+4\left(a^2+\frac{1}2\right)\ \textgreater \ 0$ . This yields two solutions $x=\dfrac{-\frac{1}{2a}\pm\sqrt{\Delta}}{2}$ , one of which is negative (and hence in the second quadrant), that is $\boxed{x=\dfrac{-\frac{1}{2a}-\sqrt{\frac{1}{4a^2}+4\left(a^2+\frac{1}2\right)}}{2}}$ .

The "extreme normal line" is the one for which that absissa is the greatest (i.e. the closest to

since they're negative). Let us this derive x(a)

with respect to

: $x'(a)=-\frac{1}2\left[-\frac{1}{a^2}+\dfrac{-\frac{8a^3}+8a}{2\sqrt{\frac{1}{4a^2}+4(a^2+\frac{1}2)}}\right]$ . $x'(a)=0\Leftrightarrow 2\sqrt{\underset{\frac{(4a^2+1)^2}{4a^2}}{\underbrace{\frac{1}{4a^2}+4(a^2+\frac{1}2)}}}+\frac{1}{8a^3}-8a=0

\Leftrightarrow 1-64a^4+8(4a^2+1)=0$ .
Setting A=a^2

, this leads to $A^2-\frac{A}2-\frac{9}{64}=0$ , whose positive root is $A=\frac{1}4+\frac{\sqrt{13}}8$ . This yields $\boxed{a=\sqrt{\frac{1}4+\frac{\sqrt{13}}8}}$ .

The equation of the extreme normal line is therefore $\boxed{y=-\frac{x}{2a}+a^2+\frac{1}2,a=\sqrt{\frac{1}4+\frac{\sqrt{13}}8}}}$

The second question is extremely similar to this one, so I'll just show you the right track :

Let's call $d(a)=\dfrac{-\frac{1}{2a}-\sqrt{\frac{1}{4a^2}+4\left(a^2+\frac{1}2\right)}}{2}$ the abcissa of the intersection of a normal ine at

and the curve, which we have computed before. The area between the normal line and the curve is $\int_{d(a)}^ax^2 dx=\left[\frac{x^3}3\right]_{d(a)}^a=\frac{1}3\left(a^3-d^3(a)\right)$ . Now, derive this with respect to

in order to find its extrema. You will some a_0

at which it is minimum, and hence the equation of the normal line that traps the least area between itself and the curve will be $y=-\frac{x}{2a_0}+a_0^2+\frac{1}2$

Don't hesitate to ask if you have any question :-) it's a very interesting problem !

Answer from: Quest

Use a calculator basically

Answer from: Quest

the largest number of the four options is 1.8 x 10-2

you have it in scientific notation.

1.80 x 10-6 = 0.00000180

1.80 x 10-4 = 0.000180 (bigger, or "less small", than the former one)

1.8 x 10-2 = 0.0180. as you see. the "less small", or the biggest one of them all.

as you have negative exponent, it means you are dividing 1.8 by 100 (i.e. moving the dot towards left, two times).

if you had positive exponent for the scientific notation, power of tenths, you would be multiplying times 100 (moving the dot towards right) as below:

1.8 x (10)2 = 180

so. if you have negative exponent in scientific notation, move the dot toward left. if it is positive, toward right as many times the exponent tells you.

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