Author Topic: Differentiation (Read 7910 times) Tweet Share

ilovevce · « **Reply #15 on:** June 20, 2009, 07:04:06 pm »

Quote from: dcc on June 20, 2009, 06:43:51 pm

Quote from: ilovevce on June 20, 2009, 06:33:13 pm
Another way to go about part (d) is:

1. Transpose the equation to make t the subject

2. Find $\frac {dt}{dm}$

3. Flip to give $\frac {dm}{dt}$

That's making it unnecessarily difficult, it'd be much easier just to do it by looking at $\frac{dm}{dt}$

Well, for this particular question, it might be easier to do it by recognition, but not in every situation. I was just giving the general rule.

d0minicz · « **Reply #16 on:** June 22, 2009, 09:20:56 pm »

If h=6, $V= 6\pi r^2 + \frac {2}{3} \pi r^3$ . For r=4:
i)Show that a small increase of p cm in the radius results in an increase of $80\pi p$ in the volume. Done
ii) Show that a small increase of q% in the radius will cause an approximate increase of 2.3%q in the volume. Need help
need to see workings not using Liebniz notation, thanks =]

d0minicz · « **Reply #17 on:** June 22, 2009, 09:52:23 pm »

Also, how do you do these q's:
A right circular cylinder is placed inside a sphere of radius 5
cm. Find the largest possible volume of the cylinder.
thanks

kamil9876 · « **Reply #18 on:** June 22, 2009, 10:11:30 pm »

first post:

ii.) $<br />p=r * \frac{q}{100}$
$=\frac{qr}{100}$

$\Delta V=\frac{4\pi qr}{5}$ (are you satisfied with that avoidance of Liebniz notation?)

Now the percentage increase of Volume is:

$\frac{\Delta V}{V(4)}*100$

ilovevce · « **Reply #19 on:** June 22, 2009, 10:24:53 pm »

Quote from: d0minicz on June 22, 2009, 09:52:23 pm

Also, how do you do these q's:
A right circular cylinder is placed inside a sphere of radius 5
cm. Find the largest possible volume of the cylinder.
thanks

Find the equation for, and graph, a circle with radius 5 centred on the origin.
Now, choose any point (x,y). The x-coordinate will be the radius of the flat surfaces of your cylinder, which will have area $A = \pi r^2$ . The y-coordinate will be half the height of you cylinder.

Now construct a formula for the volume of your cylinder in terms of x only. You can differentiate this to find the maximum value.

d0minicz · « **Reply #20 on:** June 23, 2009, 08:18:45 pm »

The tangent to the curve with equation y=tan2x at the point where $x= \frac {\pi}{8}$ meets the y=axis at point A. Find the distance OA where O is the origin.
need to check answer, thanks.

TrueTears · « **Reply #21 on:** June 23, 2009, 08:23:27 pm »

$\frac{dy}{dx} = 2sec^2(2x)$

when $x = \frac{\pi}{8}$

$\frac{dy}{dx} = 2sec^2(2(\frac{\pi}{8})) = 4$

When $x = \frac{\pi}{8}$ $y = 1$

$y - y1 = m(x-x1)$

$y - 1 = 4(x-\frac{\pi}{8})$

The distance OA is just the $|$ y intercept $|$ .

d0minicz · « **Reply #22 on:** June 23, 2009, 08:31:05 pm »

mmm the book got $\frac {\pi}{2} -1$ but i got the same as you...
thanks

Edit; dw got it now lol

julianpeiriez · « **Reply #23 on:** June 24, 2009, 06:12:34 pm »

methods cas 3 and 4 cambridge - exercise 11 C
q17

The volume V cm3, of water in a dish when the depth is "h" cm is given by the rule V = pi/2(e^2h -1). the depth of the dish is 2.5cm. If water is being poured in at 5cm^3/s, find:

a) The rate at which the depth of the water is increasing when the depth is 2cm?
plz reply asap thamx

d0minicz · « **Reply #24 on:** June 24, 2009, 06:44:55 pm »

well you find $\frac {dV}{dh}$ from the formula they give you
and we have $\frac {dV}{dt} = 5 cm/s$ and we want to find $\frac {dh}{dt}$
so, $\frac {dh}{dt} = \frac {dh}{dV}. \frac {dV}{dt}$

d0minicz · « **Reply #25 on:** June 24, 2009, 07:31:37 pm »

An aeroplane is flying horizontally at a constant height of 1000m. At a certain instant the angle of elevation is $30^o$ and decreasing and the speed of the aeroplane is 480 km/h.
a) How fast is $\theta$ decreasing at this instant? Answer is degrees/s.
b) How fast is the distance between the aeroplane and the observation point changing at this instant?
thanks

Mao · « **Reply #26 on:** June 24, 2009, 08:56:45 pm »

let x be the horizontal distance of the aeroplane from observer.

construct a triangle, through geometry, you can see that angle of elevation is opposite to vertical distance and adjacent to horizontal distance

$\tan \theta = \frac{1000}{x} \implies \theta = \tan^{-1} \frac{1000}{x}$ , plot this on the graphics calculator

Using trig knowledge, when $\theta = 30^o$ , $x=1000\sqrt{3}$ , use the graphics calculator's derivative function to find the gradient at this point, this will give you $\frac{d\theta}{dx}$

Using related rates, $\frac{d\theta}{dt} = \frac{d\theta}{dx} \cdot \frac{dx}{dt}$ , where $\frac{dx}{dt} = 480 km/hr = \frac{400}{3} m/s$

Hence, $\frac{d\theta}{dt} \approx -\frac{1}{30} radians \approx -1.91^o s^{-1}$

let y be the hypotenuse of the above triangle, this is its distance from the observer.

then by trig ratio, $y = \frac{1000}{\sin \theta}\implies \frac{dy}{d\theta} = -\frac{1000 \cos \theta}{\sin^2 \theta}$ , when $\theta = 30^o, \; \frac{dy}{d\theta} = -2000\sqrt{3}$ (be aware that $\theta$ here is in radians)

using related rates, $\frac{dy}{dt} = \frac{dy}{d\theta} \cdot \frac{d\theta}{dt} = -2000\sqrt{3} \cdot -\frac{1}{30} = \frac{200\sqrt{3}}{3} \approx 115.5 ms^{-1}$

that was an extremely tough question at methods level =\

Mao · « **Reply #27 on:** June 24, 2009, 08:58:57 pm »

Also, note to people in reply to julianpeiriez's misplacement of posts, whilst it is good to point new member to the right place, I do expect it to be done with respect and good intention. Antagonizing new members is not acceptable. If you feel it is misplaced or thread-hijacking, report the post to moderators so it can be deleted/split.

ilovevce · « **Reply #28 on:** June 26, 2009, 12:57:50 pm »

Quote from: Mao on June 24, 2009, 08:56:45 pm

$\tan \theta = \frac{1000}{x} \implies \theta = \tan^{-1} \frac{1000}{x}$ , plot this on the graphics calculator

Using trig knowledge, when $\theta = 30^o$ , $x=1000\sqrt{3}$ , use the graphics calculator's derivative function to find the gradient at this point, this will give you $\frac{d\theta}{dx}$

You can also work out $\frac{d \theta}{dx}$ without a calculator. Make $x = \frac{1000}{\tan \theta}$ and use the quotient rule to find $\frac {dx}{d \theta}$ and then flip that to get $\frac{d \theta}{dx}$ .The numbers are not too large to handle.

kamil9876 · « **Reply #29 on:** June 26, 2009, 01:29:11 pm »

Quote from: ilovevce on June 26, 2009, 12:57:50 pm

Quote from: Mao on June 24, 2009, 08:56:45 pm

$\tan \theta = \frac{1000}{x} \implies \theta = \tan^{-1} \frac{1000}{x}$ , plot this on the graphics calculator

Using trig knowledge, when $\theta = 30^o$ , $x=1000\sqrt{3}$ , use the graphics calculator's derivative function to find the gradient at this point, this will give you $\frac{d\theta}{dx}$

You can also work out $\frac{d \theta}{dx}$ without a calculator. Make $x = \frac{1000}{\tan \theta}$ and use the quotient rule to find $\frac {dx}{d \theta}$ and then flip that to get $\frac{d \theta}{dx}$ .The numbers are not too large to handle.

I think chain rule would be more efficient

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