题目

Problem

A curve has equation

\begin{align*} y=\frac{x^3}{6}+4\sqrt{x}-15,\qquad x>0. \end{align*}

(a) Find $\dfrac{dy}{dx}$ , giving the answer in simplest form.

(3)

The point $P\left(4,\dfrac{11}{3}\right)$ lies on the curve.

(b) Find the equation of the normal to the curve at $P$ . Write your answer in the form $ax+by+c=0$ , where $a$ , $b$ and $c$ are integers to be found.

(4)

解答

(a)

解法一

思路

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把 $\sqrt{x}$ 写成 $x^{1/2}$ 后逐项求导。

答题过程

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\begin{align*} y=&\,\frac16x^3+4x^{1/2}-15,\\ \frac{dy}{dx} =&\,\frac12x^2+2x^{-1/2}\\ =&\,\frac12x^2+\frac{2}{\sqrt{x}}. \end{align*}

(b)

解法一

思路

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先求切线斜率，再取负倒数得到法线斜率。最后用点斜式写直线，并整理成整数系数形式。

答题过程

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At $x=4$ ,

\begin{align*} \frac{dy}{dx} =&\,\frac12(4)^2+\frac{2}{\sqrt4}\\ =&\,8+1\\ =&\,9. \end{align*}

So the gradient of the normal is $-\frac19$ .

Using $P\left(4,\frac{11}{3}\right)$ ,

\begin{align*} y-\frac{11}{3} =&\,-\frac19(x-4). \end{align*}

Multiply by $9$ :

\begin{align*} 9y-33=&\,-x+4\\ x+9y-37=&\,0. \end{align*}

IAL 2019 Oct Q5

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程