Originally Posted by

**wintersoltice** i seriously can't differentiate between pre-calculus and calculus, so if i post in the wrong place, i'm sorry.

1. A **right** circular cone with its vertex facing down has a height of 120cm and a radius of 60cm at the top. Water is poured into the cone at a rate of $\displaystyle x$ $\displaystyle cm^3/s$. Find, in terms of x,

i) the rate at which the water level rises when the height is 60cm,

ii) the rate at which the wet inner surface of the cone is changing at this instant.

(Give your answer correct to 3 sig.fig.)

I can do 1i...the answer is 0.000354x.....1ii is the one that i don't know how to do....

**lateral surface area of a cone ...** $\displaystyle S = \pi r \sqrt{r^2+h^2}$ **... the question is asking for** $\displaystyle \frac{dS}{dt}$

2. A rectangle ABCD enclosed within the curve $\displaystyle y=8x-x^2$ and the x-axis. C and D are points on the curve A and B are points on the x-axis. If AB =2a, find the area of ABCD in terms of a. Find the value of a for which the area of the rectangle ABCD is a maximum and find this area.

I don't even know how to start.Can someone teach me how to start??

**sketch a picture of the parabola ... let point A be left of point B and note that the value **$\displaystyle a > 0$.

**point A is at position** $\displaystyle (4-a , 0)$

**point B is at position** $\displaystyle (4+a , 0)$

**rectangle height is h =** $\displaystyle 8(4-a) - (4-a)^2$

$\displaystyle A = 2a[8(4-a) - (4-a)^2]$

3. If $\displaystyle y= a cos 2x + b sin 2x +2 cos x$, show that $\displaystyle \frac{d^2y}{dx^2} + 4y$ is independent of a and b. If a and b are such that y=3 and $\displaystyle \frac{dy}{dx} =0$ when x=0, find the values of x between $\displaystyle 0^\circ$ and $\displaystyle 360^\circ$ for which $\displaystyle \frac{dy}{dx} = 0$.

i think i know how to do but my answer is not the same as the book's answer.

$\displaystyle \frac{dy}{dx} = -2a sin 2x + 2b cos 2x - 2 sin x$

$\displaystyle \frac{d^2y}{dx^2} = - 4a cos 2x - 4b sin 2x -2 cos x$

4y= 4a cos 2x + 4b sin 2x + 8 cos x

$\displaystyle \frac{d^2y}{dx^2} + 4y$

= -4a cos 2x - 4b sin 2x - 2 cos x + 4a cos 2x + 4b sin 2x + 8 cos x

=6 cos x

[shown:it's independent of a and b.]

by substituting y=3,a=2. **... I think a = 1, check that.**

by substituting $\displaystyle \frac{dy}{dx}=0$ , b=0.

-2(2)sin 2x + 2(0) cos 2x - 2 sin x = 0

-4 sin 2x - 2 sin x =0

-2(2 sin 2x + sin x)= 0

2 sin 2x + sin x =0

4 sin x cos x + sin x = 0

sin x ( 4 cos x +1) =0

sin x =0

x= 0, 180, 360

$\displaystyle cos x = (-\frac{1}{4})$

Basic angle of x = 75.5

x=104.5, 255.5

but the book's answer is 120 and 240....those in red are wrong.....where did i went wrong???