2.10.3 Differentiation SPM Practice (Paper 1 Question 21 – 30)

Question 21:

Solution:

$\begin{array}{l}\text{Let, }\\ h\text{ = height of the water level}\\ r\text{ = radius of the water surface}\\ V\text{ = volume of the water}\\ \\ \frac{r}{h}=\frac{0.4}{0.6}\leftarrow \boxed{\text{Concept of similar triangles}}\\ \frac{r}{h}=\frac{2}{3}\\ r=\frac{2}{3}h\\ \\ \text{Volume of water, }V=\frac{1}{3}\pi r^{2}h\\ V=\frac{1}{3}\pi {\left(\frac{2}{3}h\right)}^{2}h\\ V=\frac{4}{27}\pi h^3\\ \frac{dV}{dh}=\left(3\right)\frac{4}{27}\pi h^2\\ \frac{dV}{dh}=\frac{4}{9}\pi h^2\\ \\ \text{The rate of change of the height of the water level}\\ \text{when the height of water level is 0}\text{.5 }m=\frac{dh}{dt}.\\ \frac{dh}{dt}=\frac{dh}{dV}\times \frac{dV}{dt}\leftarrow \boxed{\text{Chain rule}}\\ \frac{dh}{dt}=\frac{9}{4\pi h^2}\times 0.02\leftarrow \boxed{\text{Given }\frac{dV}{dt}=0.02}\\ \frac{dh}{dt}=\frac{9}{4\pi {\left(0.5\right)}^2}\times 0.02\\ \frac{dh}{dt}=0.0572m{s}^{-1}\end{array}$

Question 22 (SPM 2019):
The curve y = px⁴ + 2x has turning point at (-1, q).
Find the value of p and of q.
[4 marks]

Answer:
$$ \begin{aligned} & y=p x^4+2 x \\ & \frac{d y}{d x}=4 p x^3+2 \\ & x=-1, \frac{d y}{d x}=-4 p+2=0 \\ & 4 p=2 \\ & p=\frac{1}{2} \end{aligned} $$


$$ \begin{aligned} (-1, q) & : x=-1, y=q \\ y & =p x^4+2 x \\ q & =\frac{1}{2}(-1)^4+2(-1) \\ = & \frac{1}{2}-2=-\frac{3}{2} \end{aligned} $$