Question: Questions from Exercise 14A
2 Let $f(x) = x^2 - 3x$. Then the points $P(4,4)$ and $Q(4 + h, (4 + h)^2 - 3(4 + h))$ are on the curve $y = f(x)$.
a Determine the gradient of the secant $PQ$.
b Determine the gradient of the tangent line to the curve at the point $P$ by considering what happens as $h$ approaches $0$.
3 The points $P(x, x^2 - 2x)$ and $Q(x + h, (x + h)^2 - 2(x + h))$ are on the curve $y = x^2 - 2x$.
Determine the gradient of $PQ$ and hence determine the derivative of $x^2 - 2x$.
6 A population of insects grows so that the size of the population, $P$, at time $t$ (days) is given by $P = 1000 + t^2 + t$. By determining the gradient of the secant through the points where $t = 3$ and $t = 3 + h$, calculate the rate of growth of the insect population at time $t = 3$.
1. **Problem 2a:** Find the gradient of the secant line $PQ$ where $P(4,4)$ and $Q(4+h, (4+h)^2 - 3(4+h))$ lie on $y = f(x) = x^2 - 3x$.
2. The gradient of the secant $PQ$ is given by the difference quotient:
$$m_{sec} = \frac{f(4+h) - f(4)}{(4+h) - 4}$$
3. Calculate $f(4+h)$:
$$f(4+h) = (4+h)^2 - 3(4+h) = (16 + 8h + h^2) - (12 + 3h) = 16 + 8h + h^2 - 12 - 3h = 4 + 5h + h^2$$
4. Calculate $f(4)$:
$$f(4) = 4^2 - 3 \times 4 = 16 - 12 = 4$$
5. Substitute into the difference quotient:
$$m_{sec} = \frac{(4 + 5h + h^2) - 4}{h} = \frac{5h + h^2}{h}$$
6. Simplify by canceling $h$:
$$m_{sec} = \frac{\cancel{h}(5 + h)}{\cancel{h}} = 5 + h$$
7. **Problem 2b:** Find the gradient of the tangent line at $P$ by letting $h \to 0$:
$$m_{tan} = \lim_{h \to 0} (5 + h) = 5$$
8. **Problem 3:** Points $P(x, x^2 - 2x)$ and $Q(x+h, (x+h)^2 - 2(x+h))$ lie on $y = x^2 - 2x$. Find the gradient of $PQ$ and the derivative.
9. Gradient of $PQ$:
$$m_{sec} = \frac{f(x+h) - f(x)}{(x+h) - x} = \frac{(x+h)^2 - 2(x+h) - (x^2 - 2x)}{h}$$
10. Expand numerator:
$$(x+h)^2 - 2(x+h) - x^2 + 2x = (x^2 + 2xh + h^2) - 2x - 2h - x^2 + 2x = 2xh + h^2 - 2h$$
11. Substitute back:
$$m_{sec} = \frac{2xh + h^2 - 2h}{h}$$
12. Simplify by canceling $h$:
$$m_{sec} = \frac{\cancel{h}(2x + h - 2)}{\cancel{h}} = 2x + h - 2$$
13. Derivative is the limit as $h \to 0$:
$$f'(x) = \lim_{h \to 0} (2x + h - 2) = 2x - 2$$
14. **Problem 6:** Population $P = 1000 + t^2 + t$. Find rate of growth at $t=3$ using secant gradient between $t=3$ and $t=3+h$.
15. Gradient of secant:
$$m_{sec} = \frac{P(3+h) - P(3)}{(3+h) - 3} = \frac{(1000 + (3+h)^2 + (3+h)) - (1000 + 3^2 + 3)}{h}$$
16. Calculate $P(3+h)$:
$$(3+h)^2 + (3+h) = (9 + 6h + h^2) + 3 + h = 12 + 7h + h^2$$
17. Calculate $P(3)$:
$$9 + 3 = 12$$
18. Substitute:
$$m_{sec} = \frac{1000 + 12 + 7h + h^2 - (1000 + 12)}{h} = \frac{7h + h^2}{h}$$
19. Simplify by canceling $h$:
$$m_{sec} = \frac{\cancel{h}(7 + h)}{\cancel{h}} = 7 + h$$
20. Rate of growth at $t=3$ is the limit as $h \to 0$:
$$\lim_{h \to 0} (7 + h) = 7$$
**Final answers:**
- 2a: Gradient of secant $PQ = 5 + h$
- 2b: Gradient of tangent at $P = 5$
- 3: Derivative of $x^2 - 2x$ is $2x - 2$
- 6: Rate of growth of population at $t=3$ is $7$