Author Topic: TyErd's questions (Read 43349 times) Tweet Share

TyErd · « **Reply #390 on:** July 02, 2010, 10:13:01 pm »

Another question, if z= sin x and sin 1=a then using linear approximation find the value of sin(1.1). I kinda get stuck half way through, can someone show me how to do it please?

brightsky · « **Reply #391 on:** July 02, 2010, 10:28:15 pm »

By linear approximation:

$f(x+h) \approx hf'(x) + f(x)$

We have $z = f(x)$ , so $f(x) = \sin(x)$ .

$\because \sin(1) = a$ ...(1)

$\therefore f(1 +0.1) \approx 0.1 \cdot \cos(1) + \sin(1)$

$f(1.1) = 0.1\cos(1) + \sin(1)$ ...(2)

From (1):

$\sin^2(1) = a^2$

$1 - \cos^2(1) = a^2$

$\cos^2(1) = 1 - a^2$

$\cos(1) = \sqrt{1 - a^2}$ (reject the negative solution)

Substitute that into (2),

$f(1.1) = 0.1 \sqrt{1-a^2} + a$

Hope this is right. :p

TyErd · « **Reply #392 on:** July 02, 2010, 10:33:55 pm »

Thankyou very much! and yes you are correct

arent you always? lol

TyErd · « **Reply #393 on:** July 02, 2010, 10:36:37 pm »

oh btw why do you have to reject the negative solution?

TyErd · « **Reply #394 on:** July 02, 2010, 11:06:20 pm »

Also how do you do this: A section of a rollercoaster is given by the equation $y= 18 cos (\frac{\pi x}{80}) + 12 , 0 \leq x \leq 80$

State the coordinates of the point of the track for which the magnitude of the gradient is maximum.

TyErd · « **Reply #395 on:** July 02, 2010, 11:12:52 pm »

Also having trouble with this one, A rocket rises vertically from level ground at a point A. It is observed from another point B on the ground where B is 10km from point A. When the angle of elevation ABR has the value $\frac {\pi}{4}$ radians, this angle is increasing at the rate of 0.005 radians per second. Find in km/s the velocity of the rocket at that instant.

I know that you have to use the chain rule but the velocity part confuses me. Any help?

brightsky · « **Reply #396 on:** July 02, 2010, 11:24:59 pm »

$\frac{dy}{dx} = -\frac{18\pi}{80} \sin\left(\frac{\pi x}{80}\right) = -\frac{9\pi}{40} \sin\left(\frac{\pi x}{80}\right)$

Treat this as another function. We want to find when the dy/dx is at its maximum. That is when $\frac{d^2y}{dx^2} = 0$ .

So $\frac{d^2y}{dx^2} = -\frac{9\pi^2}{3200} \cos \left(\frac{\pi x}{80}\right) = 0$

Simplifying this gives:

$\cos\left(\frac{\pi x}{80}\right) = 0$

$\because 0 \le x \le 80$

$\therefore 0 \le \frac{\pi x}{80} \le \pi$

$\frac{\pi x}{80} = \frac{\pi}{2}$

$x = 40$

Substituting that back into the original function:

$y = 18 \cos \left(\frac{40\pi}{80}\right) + 12 = 12$

So the point on the track for which the mangitude of the gradient is maximum is (40,12).

TyErd · « **Reply #397 on:** July 02, 2010, 11:34:50 pm »

I dont get the $\frac{d^2 y}{dx^2}$ part, how did you get that?

brightsky · « **Reply #398 on:** July 02, 2010, 11:42:52 pm »

$\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}$ where x is the distance from A to R.

We know that:

$\tan \theta = \frac{x}{10}$

$x = 10 \tan (\theta)$

$\frac{dx}{d\theta} = 10\sec^2(\theta)$ ...(1)

When $\theta = \frac{\pi}{4}$ , $\frac{dx}{d\theta} = 20$

We also know that:

$\frac{d\theta}{dt} = 0.005$ when $\theta = \frac{\pi}{4}$

Hence $\frac{dx}{dt} = 20 \times 0.005 = 0.1$ km/s.

brightsky · « **Reply #399 on:** July 02, 2010, 11:44:40 pm »

Quote from: TyErd on July 02, 2010, 11:34:50 pm

I dont get the $\frac{d^2 y}{dx^2}$ part, how did you get that?

We treat $\frac{dy}{dx} = ...$ as a normal function. To find this function is at its maximum - that is when $\frac{dy}{dx}$ is a maximum, we need to derive it and then let it = 0.

Deriving $\frac{dy}{dx}$ would yield $\frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d^2y}{dx^2}$

brightsky · « **Reply #400 on:** July 02, 2010, 11:51:57 pm »

Quote from: TyErd on July 02, 2010, 10:36:37 pm

oh btw why do you have to reject the negative solution?

Hehe, took a shortcut but we know $\cos(1)$ is positive given its graph. It's only between $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ that the graph goes negative.

TyErd · « **Reply #401 on:** July 02, 2010, 11:59:36 pm »

Quote from: brightsky on July 02, 2010, 11:44:40 pm

Quote from: TyErd on July 02, 2010, 11:34:50 pm
I dont get the $\frac{d^2 y}{dx^2}$ part, how did you get that?

We treat $\frac{dy}{dx} = ...$ as a normal function. To find this function is at its maximum - that is when $\frac{dy}{dx}$ is a maximum, we need to derive it and then let it = 0.

Deriving $\frac{dy}{dx}$ would yield $\frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d^2y}{dx^2}$

Oh okay so because gradient is dy/dx, where finding when that is maximum so as usual we differentiate and make it equal zero. Okay I get it thanks

brightsky · « **Reply #402 on:** July 03, 2010, 12:00:42 am »

Quote from: TyErd on July 02, 2010, 11:59:36 pm

Oh okay so because gradient is dy/dx, where finding when that is maximum so as usual we differentiate and make it equal zero.

Yep.

TyErd · « **Reply #403 on:** July 03, 2010, 12:06:24 am »

Thanks so much brightsky! you've helped me so much today really appreciate it!

TyErd · « **Reply #404 on:** July 03, 2010, 12:08:54 am »

Um what do you need to know about the fundamental theorem of calculus? There's like 3 pages explaining it but is it important to know?

Search the forums now!

We have moved!

Author Topic: TyErd's questions (Read 43349 times) Tweet Share

TyErd

Re: TyErd's questions

brightsky

Re: TyErd's questions

TyErd

Re: TyErd's questions

TyErd

Re: TyErd's questions

TyErd

Re: TyErd's questions

TyErd

Re: TyErd's questions

brightsky

Re: TyErd's questions

TyErd

Re: TyErd's questions

brightsky

Re: TyErd's questions

brightsky

Re: TyErd's questions

brightsky

Re: TyErd's questions

TyErd

Re: TyErd's questions

brightsky

Re: TyErd's questions

TyErd

Re: TyErd's questions

TyErd

Re: TyErd's questions

Recent Posts

User:
Password: