1. differentiation

1. the curve y=ax^2+bx passes through the point (2,4) with gradient 8.

find a and b

2. show that if y=2x-x^2 then

y 2^2y - 2 dy +2y = 4(x+1)
dx^2 dx

3. A child's height h cm at age 8 years can be modelled by the equation

h= _ a^4
500 + (a-1/2)^2 +55

for ages 11 ≤ a ≤ 16

Find the child's annual growth rate at age 12, and at age 15

I am completely stuck with what to do, so a little guidance with how to go about solving would be great. Thanks

2. Hello, lra11!

1. The curve $y\:=\:ax^2+bx$ passes through the point $(2,4)$ with gradient $8.$

Find $a$ and $b.$

We have: . $y \:=\:ax^2+b$

Since (2,4) is on the curve: . $4 \:=\:a\!\cdot\!2^2 + b\quad\Rightarrow\quad 4a + b \:=\:4$ .[1]

The gradient is given by: . $y' \:=\:2ax$
. . At (2,4), we have: . $8 \:=\:2a(2)\quad\Rightarrow\quad\boxed{a \,=\,2}$

Substitute into [1]: . $4(2) + b \:=\:4\quad\Rightarrow\quad\boxed{b \:=\:-4}$

2. Show that if $y\:=\:2x-x^2$, then

y 2^2y - 2 dy +2y = 4(x+1) . . . . . impossible to read this!
dx^2 dx
If I were to guess what it said . . . maybe: . $y\!\cdot\!\frac{d^2y}{dx^2} - 2\!\cdot\!\frac{dy}{dx} + 2y \;=\;4(x+1)$
. . but it is not true!

We have: . $\begin{Bmatrix}y & = & 2x-x^2 \\ \\ \dfrac{dt}{dx} & = & 2-2x \\ \\ \dfrac{d^2y}{dx^2} &=& -2 \end{Bmatrix}$

Then: . $\underbrace{y}_{\downarrow}\;\;\cdot\;\;\underbrac e{\frac{d^2y}{dx^2}}_{\downarrow} \:- \:2 \cdot \underbrace{\frac{dy}{dx}}_{\downarrow} \;\;\;+\;\;\;2\;\cdot\;\underbrace{y}_{\downarrow}$
. . . . . $(2x-x^2)(-2) - 2(2-2x) + 2(2x-x^2)$

. . . . $-4x + 2x^2 - 4 + 4x + 4x - 2x^2$

. . . $= \;4x - 4 \;=\;{\color{blue}4(x-1)}$

3. A child's height $h$ cm at age $a$ years can be modelled by the equation

. . $h \;=\;-\frac{a^4}{500} + \left(a - \frac{1}{2}\right)^2 + 55$ . for ages $11 \leq a \leq 16$ . . . . another guess

Find the child's annual growth rate at age 12, and at age 15.

Differentiate: . $\frac{dh}{da} \;=\;-\frac{4a^3}{500} + 2\left(a - \frac{1}{2}\right) \;=\;-\frac{a^3}{125} + 2a - 1$

When $a = 12\!:\;\;\frac{dh}{da} \;=\;-\frac{12^3}{125} + 2(12) - 1 \;=\;9.176\text{ cm/year}$

When $a = 15\!:\;\;\frac{dh}{dt} \;=\;-\frac{15^3}{125} + 2(15) - 1 \;=\;2\text{ cm/year}$