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CIE 9231 2023 November Paper 22 Q2

A Level / CIE / FP2

CIE 9231 2023 Nov Paper 22 Paper · Question 2

题目

Problem

It is given that

x=1+1tandy=tet.x=1+\frac{1}{t}\qquad\text{and}\qquad y=te^t.

(a) Show that

dydx=−et(t3+t2).\frac{\mathrm{d}y}{\mathrm{d}x}=-e^t\big(t^3+t^2\big).
[3]

(b) Find d2ydx2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} in terms of tt.

[4]
题目中文翻译

已知

x=1+1t以及y=tetx=1+\frac{1}{t}\qquad\text{以及}\qquad y=te^t

(a) 证明

dydx=−et(t3+t2)\frac{\mathrm{d}y}{\mathrm{d}x}=-e^t\big(t^3+t^2\big)
(3)

(b) 求用 tt 表示的 d2ydx2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}。

(4)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

分别对 x(t),y(t)x(t),y(t) 求导,再用链式法则取比值。

答题过程

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The parameter requires t≠0t\ne0. The derivatives are dxdt=−t−2\frac{\mathrm{d}x}{\mathrm{d}t}=-t^{-2} and dydt=(t+1)et\frac{\mathrm{d}y}{\mathrm{d}t}=(t+1)e^t. Hence

dydx=−t2(t+1)et=−et(t3+t2).\frac{\mathrm{d}y}{\mathrm{d}x}=-t^2(t+1)e^t=-e^t(t^3+t^2).

(b)

解法一

思路

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对 (a) 用乘积法则求关于 tt 的导数,再乘 dtdx=−t2\frac{\mathrm{d}t}{\mathrm{d}x}=-t^2。

答题过程

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Differentiating with respect to tt gives

ddt(dydx)= −et(t3+t2)−et(3t2+2t)= −tet(t2+4t+2).\begin{align*} \frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg) =&\,-e^t(t^3+t^2)-e^t(3t^2+2t)\\ =&\,-te^t(t^2+4t+2). \end{align*}

Multiplying by −t2-t^2 gives d2ydx2=t3et(t2+4t+2)\boxed{\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=t^3e^t(t^2+4t+2)}.