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A-Level Pure 4 · Edexcel IAL WMA14 · Topic pack

Topic pack · Implicit differentiation & rates of change แพ็กฝึกเฉพาะหัวข้อ · อิมพลิซิต · อัตราการเปลี่ยนแปลง

10 questions 10 ข้อ 55 marks 55 คะแนน 70 min 70 นาที Implicit differentiation & connected rates, easy → hard

Ten questions on one topic, ordered easy → hard. The last three are full exam difficulty. Open the mark schemes only after a real attempt. สิบข้อหัวข้อเดียว เรียงง่าย → ยาก สามข้อสุดท้ายคือระดับข้อสอบจริง เปิดมาร์คสกีมหลังลองทำจริงเท่านั้น

⏱ 70:00 Scoreคะแนนรวม: –
Q1. 3 marks

A curve has equation $x^2+y^2=25$. Find the gradient of the curve at the point $(3,4)$.

Mark schemeshow ▾
M1 differentiate implicitly: $2x+2y\dfrac{dy}{dx}=0$  M1 rearrange: $\dfrac{dy}{dx}=-\dfrac{x}{y}$  A1 at $(3,4)$: gradient $=-\dfrac34$.
Q2. 4 marks

The curve $C$ has equation $x^2+xy=12$. Find the gradient of $C$ at the point $(2,4)$.

Mark schemeshow ▾
B1 product rule on $xy$: $\dfrac{d}{dx}(xy)=y+x\dfrac{dy}{dx}$  M1 $2x+y+x\dfrac{dy}{dx}=0$  M1 $\dfrac{dy}{dx}=-\dfrac{2x+y}{x}$  A1 at $(2,4)$: $-\dfrac{4+4}{2}=-4$.
Q3. 4 marks

Air is pumped into a spherical balloon so that its volume increases at a constant rate of $100\text{ cm}^3\text{ s}^{-1}$. Find the rate of increase of the radius, in $\text{cm s}^{-1}$, at the instant when the radius is $5$ cm. Give an exact answer. $\left[V=\tfrac43\pi r^3\right]$

Mark schemeshow ▾
B1 $\dfrac{dV}{dr}=4\pi r^2$  M1 chain rule: $\dfrac{dr}{dt}=\dfrac{dV}{dt}\div\dfrac{dV}{dr}=\dfrac{100}{4\pi r^2}$  M1 substitute $r=5$: $\dfrac{100}{100\pi}$  A1 $\dfrac{dr}{dt}=\dfrac{1}{\pi}\text{ cm s}^{-1}$.
Q4. 5 marks

The curve $C$ has equation $x^3+y^3=9xy$. Show that $\dfrac{dy}{dx}=\dfrac{3y-x^2}{y^2-3x}$, and hence find the gradient of $C$ at the point $(2,4)$.

Mark schemeshow ▾
M1 $3x^2+3y^2\dfrac{dy}{dx}=\ldots$  B1 product rule on $9xy$: $9y+9x\dfrac{dy}{dx}$  M1 collect: $(3y^2-9x)\dfrac{dy}{dx}=9y-3x^2$  A1 $\dfrac{dy}{dx}=\dfrac{3y-x^2}{y^2-3x}$ ∎  A1 at $(2,4)$: $\dfrac{12-4}{16-6}=\dfrac{4}{5}$.
Q5. 5 marks

A circular oil patch expands so that its area increases at a constant rate of $8\text{ cm}^2\text{ s}^{-1}$. At the instant when the radius is $4$ cm, find (a) the rate of increase of the radius, (b) the rate of increase of the circumference. Give exact answers.

Mark schemeshow ▾
(a) B1 $A=\pi r^2 \Rightarrow \dfrac{dA}{dr}=2\pi r$  M1 $\dfrac{dr}{dt}=\dfrac{dA}{dt}\div\dfrac{dA}{dr}=\dfrac{8}{8\pi}$  A1 $=\dfrac{1}{\pi}\text{ cm s}^{-1}$.
(b) M1 $C=2\pi r \Rightarrow \dfrac{dC}{dt}=2\pi\dfrac{dr}{dt}=2\pi\cdot\dfrac{1}{\pi}$  A1 $=2\text{ cm s}^{-1}$.
Q6. 6 marks

The curve $C$ has equation $x^2+xy+y^2=7$. Find an equation of the tangent to $C$ at the point $(1,2)$, giving your answer in the form $ax+by=c$ where $a$, $b$ and $c$ are integers.

Mark schemeshow ▾
M1 differentiate implicitly  B1 product rule: $\dfrac{d}{dx}(xy)=y+x\dfrac{dy}{dx}$, giving $2x+y+(x+2y)\dfrac{dy}{dx}=0$  M1 $\dfrac{dy}{dx}=-\dfrac{2x+y}{x+2y}$  A1 at $(1,2)$: $-\dfrac{4}{5}$  M1 $y-2=-\dfrac45(x-1)$  A1 $4x+5y=14$.
Q7. 6 marks

Water is poured into an empty container in the shape of an inverted cone at a constant rate of $18\text{ cm}^3\text{ s}^{-1}$. At every instant, the radius $r$ of the water surface and the depth $h$ of the water satisfy $r=\dfrac{h}{2}$. (a) Show that the volume of water is $V=\dfrac{\pi h^3}{12}$. (b) Find, in exact form, the rate at which the depth is increasing when $h=6$ cm.

Mark schemeshow ▾
(a) M1 $V=\dfrac13\pi r^2 h$ with $r=\dfrac h2$: $V=\dfrac13\pi\dfrac{h^2}{4}h$  A1 $V=\dfrac{\pi h^3}{12}$ ∎.
(b) B1 $\dfrac{dV}{dh}=\dfrac{\pi h^2}{4}$  M1 $\dfrac{dh}{dt}=\dfrac{dV}{dt}\div\dfrac{dV}{dh}=\dfrac{18}{\pi h^2/4}$  M1 at $h=6$: $\dfrac{18}{9\pi}$  A1 $\dfrac{dh}{dt}=\dfrac{2}{\pi}\text{ cm s}^{-1}$.
Q8. 7 marks

The curve $C$ has equation $x^3-2xy+y^3=5$. The point $P(2,1)$ lies on $C$. (a) Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$, and show that the gradient of $C$ at $P$ is $10$. (b) Find an equation of the normal to $C$ at $P$, giving your answer in the form $ax+by=c$ where $a$, $b$ and $c$ are integers.

Mark schemeshow ▾
(a) M1 $3x^2+3y^2\dfrac{dy}{dx}=\ldots$  B1 product rule: $\dfrac{d}{dx}(2xy)=2y+2x\dfrac{dy}{dx}$  M1 collect: $\dfrac{dy}{dx}=\dfrac{2y-3x^2}{3y^2-2x}$  A1 at $(2,1)$: $\dfrac{2-12}{3-4}=10$ ∎.
(b) M1 normal gradient $=-\dfrac{1}{10}$  M1 $y-1=-\dfrac{1}{10}(x-2)$  A1 $x+10y=12$.
Q9. 7 marks

The curve $C$ has equation $x^2+xy+y^2=27$. Find the coordinates of the two points on $C$ at which the tangent is parallel to the $x$-axis.

Mark schemeshow ▾
M1 differentiate implicitly: $2x+y+(x+2y)\dfrac{dy}{dx}=0$  A1 $\dfrac{dy}{dx}=-\dfrac{2x+y}{x+2y}$  M1 horizontal tangent ⇒ numerator $=0$  A1 $y=-2x$  M1 substitute into curve: $x^2-2x^2+4x^2=27$, so $3x^2=27$  A1 $x=\pm 3$  A1 points $(3,-6)$ and $(-3,6)$.
Q10. 8 marks

A spherical balloon is inflated so that its volume increases at a constant rate of $200\text{ cm}^3\text{ s}^{-1}$. $\left[V=\tfrac43\pi r^3,\ S=4\pi r^2\right]$ (a) Find the rate of increase of the radius when $r=5$ cm. (b) Find the rate of increase of the surface area at the same instant. (c) Find the radius at the instant when the radius is increasing at $\dfrac{1}{2\pi}\text{ cm s}^{-1}$. Give exact answers.

Mark schemeshow ▾
(a) B1 $\dfrac{dV}{dr}=4\pi r^2$  M1 $\dfrac{dr}{dt}=\dfrac{200}{4\pi r^2}=\dfrac{200}{100\pi}$  A1 $=\dfrac{2}{\pi}\text{ cm s}^{-1}$.
(b) B1 $\dfrac{dS}{dr}=8\pi r$  M1 $\dfrac{dS}{dt}=8\pi r\cdot\dfrac{dr}{dt}=40\pi\cdot\dfrac{2}{\pi}$  A1 $=80\text{ cm}^2\text{ s}^{-1}$.
(c) M1 $\dfrac{200}{4\pi r^2}=\dfrac{1}{2\pi}\Rightarrow r^2=100$  A1 $r=10$ cm.