Capriccio of stars in Mathematics · Skill volume

The second derivative of composite function and the solution of parametric equation curvature

The second derivative of a composite function

Derivation algorithm through function multiplication , After calculation, the result is ：
$$\begin{array}{rl}& y=y(x),x=x(t)\\ & \\ & x^{\prime }(t)=\frac{dx}{dt},y^{\prime }(t)=\frac{dy}{dt},x^{\prime \prime }(t)=\frac{d^{2}x}{d{t}^2},y^{\prime \prime }(t)=\frac{d^{2}y}{d{t}^2}\\ & \\ & \frac{d^{2}y}{d{x}^2}=[]\cdot y^{\prime \prime }(t)-[]\cdot x^{\prime \prime }(t)\\ & \\ & \frac{d^{2}y}{d{x}^2}=[\frac{x^{\prime }(t)}{x^{\prime }(t{)}^3}]\cdot y^{\prime \prime }(t)-[\frac{y^{\prime }(t)}{x^{\prime }(t{)}^3}]\cdot x^{\prime \prime }(t)\\ & \frac{d^{2}y}{d{x}^2}=[\frac{x^{\prime }(t)}{x^{\prime }(t{)}^3}]\cdot y^{\prime \prime }(t)-[\frac{y^{\prime }(t)}{x^{\prime }(t{)}^3}]\cdot x^{\prime \prime }(t)=\frac{\left|\begin{array}{ll}{x}^{\prime }(t)& x^{\prime \prime }(t)\\ y^{\prime }(t)& y^{\prime \prime }(t)\end{array}\right|}{x^{\prime }(t{)}^3}\end{array}$$

The parametric equation solves the curvature

The curvature formula is ： $k=\left|\frac{y^{\prime \prime }}{{\left(1+y^{\prime 2}\right)}^{\frac{3}{2}}}\right|$ , It means , The core demand of solving curvature is to solve the value of first-order derivative and second-order derivative .

Because in the process of finding the second derivative of the composite function , We calculate the second derivative by taking $x$,$y$ Regard them as $t$ Function of , And this also exactly conforms to the form of parametric equation , therefore , For the second derivative of a parametric equation , We still have something to do with 【 The second derivative of a composite function 】 The same conclusion .

Then substitute the curvature formula , After calculation and simplification, we can get ：
$$\begin{array}{rl}& y=y(x),x=x(t)\\ & \\ & x^{\prime }(t)=\frac{dx}{dt},y^{\prime }(t)=\frac{dy}{dt},x^{\prime \prime }(t)=\frac{d^{2}x}{d{t}^2},y^{\prime \prime }(t)=\frac{d^{2}y}{d{t}^2}\\ & \frac{d^{2}y}{d{x}^2}=[\frac{x^{\prime }(t)}{x^{\prime }(t{)}^3}]\cdot y^{\prime \prime }(t)-[\frac{y^{\prime }(t)}{x^{\prime }(t{)}^3}]\cdot x^{\prime \prime }(t)=\frac{\left|\begin{array}{ll}{x}^{\prime }(t)& x^{\prime \prime }(t)\\ y^{\prime }(t)& y^{\prime \prime }(t)\end{array}\right|}{x^{\prime }(t{)}^3}\\ & \\ & k=\frac{|x^{\prime }(t)y^{\prime \prime }(t)-x^{\prime \prime }(t)y^{\prime }(t)|}{(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2{)}^{\frac{3}{2}}}=\frac{\left|\left|\begin{array}{ll}{x}^{\prime }(t)& x^{\prime \prime }(t)\\ y^{\prime }(t)& y^{\prime \prime }(t)\end{array}\right|\right|}{(\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}{)}^3}\end{array}$$
When substituting into the equation , We can solve the curvature with the help of tabular parametric equation （ Means of substituting values ）,

follow 【 Cross multiply and subtract , Square sum open root sign 】 Principles ,

Take the title as an example ：
$$curve\{\begin{array}{l}x=t^2+2t\\ y=3\ln t\end{array}The\; upper\; corresponds\; tot=1The\; curvature\; at\; the\; point\; of\; is$$
The answer process ：
$$k=\frac{\left|\left|\begin{array}{ll}{x}^{\prime }(t)& x^{\prime \prime }(t)\\ y^{\prime }(t)& y^{\prime \prime }(t)\end{array}\right|\right|}{(\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}{)}^3}=\frac{\left|\left|\begin{array}{ll}& \\ & \end{array}\right|\right|}{(\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}{)}^3}(Just\; write\; this\; on\; draft\; paper)$$

$$\begin{array}{rl}& \left|\begin{array}{ll}4& 2\\ 3& -3\end{array}\right|=-18\to 18\\ & \\ & \sqrt{4^2+3^2}=5\to 5^3=125\\ & \\ & k=\frac{18}{125}\end{array}$$

The answer is over .

Thank you very much ： Formula derivation and tabular method are inspired by $CCL$ .