Question 2:
Given the equation of a curve is:
y = x² (x – 3) + 1
(a) Find the gradient of the curve at the point where x = –1.
(b) Find the coordinates of the turning points.

Solution:
(a)
$\begin{array}{l}y=x^2\left(x-3\right)+1\\ y=x^3-3x^2+1\\ \frac{dy}{dx}=3x^2-6x\\ \text{When }x=-1\\ \frac{dy}{dx}=3{\left(-1\right)}^2-6\left(-1\right)\\ =9\\ \text{Gradient of the curve is 9}\text{.}\end{array}$

(b)
At turning points, $\frac{dy}{dx}=0$
3x² – 6x = 0
x² – 2x = 0
x (x – 2) = 0
x = 0, 2

y = x² (x – 3) + 1
When x = 0, y = 1
When x = 2,
y = 2² (2 – 3) + 1
y = 4 (–1) + 1 = –3
Therefore, coordinates of the turning points are (0, 1) and (2, –3).

Question 3:
It is given the equation of the curve is y = 2x (1 – x)⁴ and the curve pass through T(2, 4).
Find
(a) the gradient of the curve at point T.
(b) the equation of the normal to the curve at point T.

Solution:
(a)
$\begin{array}{l}y=2x{\left(1-x\right)}^4\\ \frac{dy}{dx}=2x\times 4{\left(1-x\right)}^3\left(-1\right)+{\left(1-x\right)}^4\times 2\\ =-8x{\left(1-x\right)}^3+2{\left(1-x\right)}^4\\ \\ \text{At }T\left(2,4\right),x=2.\\ \frac{dy}{dx}=-16\left(-1\right)+2\left(1\right)\\ =16+2\\ =18\end{array}$

(b)
$\begin{array}{l}\text{Equation of normal:}\\ y-y_1=-\frac{1}{\frac{dy}{dx}}\left(x-x_1\right)\\ y-4=-\frac{1}{18}\left(x-2\right)\\ 18y-72=-x+2\\ x+18y=74\end{array}$

Question 1:

Solution:

$\begin{array}{l}\text{when }x=1,\\ \frac{d^{2}y}{d{x}^2}=6x-12\\ \frac{d^{2}y}{d{x}^2}=6\left(1\right)-12\\ \frac{d^{2}y}{d{x}^2}=-6<0\\ \text{Since }\frac{d^{2}y}{d{x}^2}<0,B\text{ is a maximum point}\text{.}\end{array}$

Question 23:
Solution:
(a)
$\begin{array}{l}y=15x+\frac{24}{x^3}\\ y=15x+24x^{-3}\\ \frac{dy}{dx}=15-72x^{-4}\\ \frac{dy}{dx}=15-\frac{72}{x^4}\\ \\ \text{When }x=2\\ \frac{dy}{dx}=15-\frac{72}{2^4}=10.5\end{array}$

(b)
$\begin{array}{l}\text{Approximate change in }y\text{ to }x\text{ in terms of }k,\\ \frac{\delta y}{\delta x}\approx \frac{dy}{dx}\\ \delta y=\frac{dy}{dx}\times \delta x\\ \delta y=10.5\times \left(2+k-2\right)\\ \delta y=10.5k\end{array}$

Question 24:
Solution:
$\begin{array}{l}\text{Area of circle, }A=\pi r^2\\ \frac{dA}{dr}=2\pi r\\ \\ \text{Approximate increase in the area to radius,}\\ \frac{\delta A}{\delta r}\approx \frac{dA}{dr}\\ \delta A=\frac{dA}{dr}\times \delta r\\ \delta A=\left(2\pi r\right)\times \left(4.01-4\right)\\ \delta A=\left[2\pi \left(4\right)\right]\times \left(0.01\right)\\ \delta A=0.08\pi {\text{ cm}}^2\end{array}$

Question 25:
(b) If x increases from 5 to 5.01, find the small increase in t.
Solution:
$\begin{array}{l}y=3t+5t^2\\ \frac{dy}{dt}=3+10t\\ \\ x=5t-1\\ \frac{dx}{dt}=5\end{array}$

(a)
$\begin{array}{l}\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}\\ \frac{dy}{dx}=\left(3+10t\right)\times \frac{1}{5}\\ \frac{dy}{dx}=\frac{3+10\left(\frac{x+1}{5}\right)}{5}\leftarrow \boxed{\begin{array}{l}x=5t-1\\ t=\frac{x+1}{5}\end{array}}\\ \frac{dy}{dx}=\frac{3+2x+2}{5}\\ \frac{dy}{dx}=\frac{5+2x}{5}\end{array}$

(b)
$\begin{array}{l}\text{Small increase in }t\text{ to }x,\\ \delta t=\frac{dt}{dx}\times \delta x\\ \delta t=\frac{1}{5}\times \left(5.01-5\right)\\ \delta t=0.002\end{array}$

Question 20:

Solution:
$\begin{array}{l}V=\frac{1}{5}{h}^3+7h\\ \frac{dV}{dh}=\frac{3}{5}{h}^2+7=\frac{3h^2+35}{5}\\ \\ \text{Given }\frac{dV}{dt}=15\text{, }h=3\\ \\ \text{Rate of change of the height of water}=\frac{dh}{dt}\\ \frac{dh}{dt}\text{=}\frac{dh}{dV}\times \frac{dV}{dt}\leftarrow \boxed{\text{Chain rule}}\\ \\ \frac{dh}{dt}\text{=}\frac{5}{3h^2+35}\times 15\\ \\ \frac{dh}{dt}\text{=}\frac{75}{62}{\text{ cms}}^{-1}\end{array}$

Question 21:
Solution:
$\begin{array}{l}\text{Length of circumference of a circle, }\\ L=2\pi r\\ \frac{dL}{dr}=2\pi \end{array}$

(a)
$\begin{array}{l}\text{Given }\frac{dL}{dt}=0.3\\ \text{Rate of change in the radius of the circle}=\frac{dr}{dt}\\ \frac{dr}{dt}=\frac{dr}{dL}\times \frac{dL}{dt}\\ \frac{dr}{dt}=\frac{1}{2\pi }\times 0.3\\ \frac{dr}{dt}=0.0477{\text{ cms}}^{-1}\end{array}$

(b)
$\begin{array}{l}2\pi r=88\\ r=\frac{88}{2\pi }=\frac{44}{\pi }\\ \\ \text{Hence, the radius of the circle after }5s\\ =\frac{44}{\pi }+5\left(0.0477\right)\\ =14.24\text{ cm}\end{array}$

Question 22:
Solution:

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