I m the one who asked you last time about the question about differentiation, and now I have a quest

Hello, I haven't come to do the question for the past two days, so I've kept you waiting for a long time.

Curves and shapes in a Cartesian coordinate system are transformed orthogonally (translated and rotated) without changing their shape and size.

The parabola can always be orthogonally transformed into a graph with the opening pointing up and the vertex at the origin, and its analytic formula is .

y=ax 2 (a>0), it doesn't hurt that the ends of one of its strings are m(m,am 2),n(n,an 2), and m is the straight line mn is y-am 2=[a(n 2-m 2) (n-m)](x-m), i.e., y=a(m+n)x-amn

The tangents of m and n are y-am 2=2am(x-m)+am 2=2amx-am 2, y=2anx-an 2, and the intersection point of the two tangents is obtained by simultaneous solution, p((m+n) 2, amn)).

The ratio of the area s1 enclosed by the parabola and one of its strings to the area s2 of the triangle mpn is .

s1/s2=∫(m,n)[a(m+n)x-amn-ax^2]dx / [ m,（m+n）/2) (a(m+n)x-amn-2amx+am^2)dx +∫m+n）/2,n) (a(n+m)x-amn-2anx+an^2) dx]

n-m)^3/6]/ [ m,（m+n）/2) (n-m)x-mn+m^2)dx+∫(m+n）/2,n) (m-n)x-mn+n^2) dx]

n-m)^3/6]/ [ n-m)^3 /8+(n-m)^3 /8]=2/3

Note: You can also use a parabola with an opening pointing downwards in this problem, the chord is below when calculating the area, and the parabola and tangent are above, and the integrand is subtracted from the former by the integrand, and the result is the same. Of course, straight-edged graphs can also use the coordinates of points to calculate trapezoidal and triangular areas.

I'm not that person.

First of all, take the vertex of the parabola as the origin point to make the plane Cartesian coordinate system, so that the equation of the parabola is y=ax 2

Take a(x1,ax1 2)b(x2,ax2 2) on the parabola and make perpendicular lines of the x-axis through a and b respectively to obtain a trapezoid.

The area enclosed by the parabola and one of its strings s1 = trapezoidal area - the area of the parabola and the x-axis and the two perpendiculars.

1 2)*(x1-x2)'s absolute value *(ax1 2+ax2 2)-

1/6）*a*（|x2-x1|）^3

The area of the triangle formed by the two tangents at both ends of this string and the chord = s2 The two vertices of this triangle are a and b, let the remaining one be c through the tangent of a and b ya=2ax1x-ax1 2 yb=2ax2x-ax2 2, and the simultaneous ya, yb

Get c((x1+x2) 2,ax1x2) Here, all three coordinates are available.

So s2=(1 2)*|x1 ax1^2 1|

x2 ax2^2 1|

x1+x2）/2 ax1x2 1|

1/4）*a*（|x2-x1|）^3

i.e. s1 s2=(1 6) (1 4)=4 6=2 3.

δy=a(x)δx+ (x) The former is the linear part and the latter is the higher order infinitesimal quantity. Here the point x is a constant, and a(x) is also a constant, so when δx tends to 0, a(x)δx and δx are infinitesimal of the same order, and oδx=δy-a(x)δx is an infinitesimal of a higher order than δx. Note that δx is not an arbitrary value, and oδx will be smaller than δx only if the absolute value of δx is less than a certain value (which may be very small).

For example: y= x, x=1, a(x)=1 2, when δx=, δy= , a(x)δx=, (x)=

As δx becomes smaller, (x) becomes even more insignificant. Again, a(x)δx and (δx) are compared as δx approaches 0. In differential triangles, δx should be studied when it is small enough.

I guess that's understandable. I wish you all the best in your exploration!