Author Topic: VCE Methods Question Thread! (Read 5782416 times) Tweet Share

Conic · « **Reply #5340 on:** July 12, 2014, 01:53:31 pm »

Quote from: Anchy on July 10, 2014, 10:09:10 pm

An isosceles triangle has its equal sides of length 10cm with an included angle (theta). If theta changes from 60degrees to 61 degrees, find correct to 2 decimal places:

1. The approx area of triangle when theta=61degrees
2. The approximate increase in the area, A of a triangle

Thanks

For a triangle with an angle of $\theta$ between sides of length a and b, the area is given by $\frac{1}{2}\cdot a\cdot b \cdot \sin(\theta)$ , so we have

$A(61^{\circ})=\frac{1}{2}\cdot10\cdot10\cdot\sin(61^{\circ})=50\sin(61^{\circ})\approx 43.73.$

For the next part, we use linear approximation. The change in theta is $1^{\circ}$ , which is $\frac{\pi}{180}$ , so this is the h value.

$A(\theta)=50\sin(\theta) \implies A'(\theta)=50\cos(\theta).$

Now using the formula for linear approximation,

$A(61^{\circ})\approx A(60^{\circ})+h\:A'(60^{\circ})$

$\approx 50\sin(60^{\circ})+\frac{\pi}{180}\cdot 50\cos(60^{\circ}).$

The increase is the final area minus the initial (the initial is $50\sin(60^{\circ})$ ), so the change is

$\Delta A \approx \frac{\pi}{180}\cdot 50\cos(60^{\circ}).$

$\approx 0.44.$

soNasty · « **Reply #5341 on:** July 12, 2014, 02:13:37 pm »

Using Z~N(0,1) find:
Pr(|Z|<k)=0.85

keltingmeith · « **Reply #5342 on:** July 12, 2014, 02:19:11 pm »

By the symmetry of the standard normal distribution:
$Pr(|Z|<k)=Pr(-k<Z<k) = 2(0.5 - Pr(Z< -k)) = 0.85<br />\\ \therefore Pr(Z<-k)=0.075$

You can then find k using the inverse normal function on your calculator.

TrueTears · « **Reply #5343 on:** July 12, 2014, 03:33:50 pm »

Quote from: soNasty on July 12, 2014, 02:13:37 pm

Using Z~N(0,1) find:
Pr(|Z|<k)=0.85

$k=\sqrt{\pi\,\mbox{erf}^{-1}(0.85)}$ where $\text{erf}$ is the error function.

Anchy · « **Reply #5344 on:** July 12, 2014, 05:37:00 pm »

Find the radius of a 1 litre cylindrical can, which will minimize the cost of metal to make it

Thanks

Rishi97 · « **Reply #5345 on:** July 12, 2014, 06:32:41 pm »

Quote from: Anchy on July 12, 2014, 05:37:00 pm

Find the radius of a 1 litre cylindrical can, which will minimize the cost of metal to make it

Thanks

Is that all the info that is given?
I was thinking of finding the derivative but I don't think there is enough info to do that.

Orb · « **Reply #5346 on:** July 12, 2014, 06:53:18 pm »

Quote from: Rishi97 on July 12, 2014, 06:32:41 pm

Is that all the info that is given?
I was thinking of finding the derivative but I don't think there is enough info to do that.

Well there actually is.
Surface area of cylinder = 2(pi)r^2 +2(pi)rh

We want to find the minimum surface area possible so we differentiate the above equation with respect to r.

= 4(pi)r + 2(pi)h
Solve it for 0, but before we do that we need to find h in terms of r

The total volume of the cylinder is 1L, the volume equation is V= pi r^2h
So (pi)r^2h=1000
re-arranging it, we get h=(1000/{(pi)r^2})

Sub ^ into 4(pi)r+2(pi)h, so
4(pi)r+2000/r^2 = 0

Solve it for r, and you get r= cube root (500/pi)

AngelWings · « **Reply #5347 on:** July 13, 2014, 12:32:09 pm »

Please help me with this question. I know there's something I'm missing here that should be obvious.

Thanks in advance.

jessss0407 · « **Reply #5348 on:** July 13, 2014, 06:53:07 pm »

Hey guys! I have a question..

Find the derivative of y=5x cos(2x)
hence use your result to find the exact value of $\int_{0}^{pi/4} (-2xsin2x)dx$

Thanks!

Zealous · « **Reply #5349 on:** July 13, 2014, 07:13:11 pm »

Quote from: jessss0407 on July 13, 2014, 06:53:07 pm

Hey guys! I have a question..

Find the derivative of y=5x cos(2x)
hence use your result to find the exact value of $\int_{0}^{pi/4} (-2xsin2x)dx$

Thanks!

Integration by recognition:

$\begin{eqnarray*}\\y & = & 5x\cos(2x)\\\frac{dy}{dx} & = & 5\cos(2x)-10x\sin(2x)\\\\\int(5\cos(2x)-10x\sin(2x))dx & = & 5x\cos(2x)\\\int5\cos(2x)dx+\int-10x\sin(2x)dx & = & 5x\cos(2x)\\\int-10x\sin(2x)dx & = & 5x\cos(2x)-\int5\cos(2x)dx\\2\int-5x\sin(2x)dx & = & 5x\cos(2x)-\frac{5}{2}\sin(2x)\\\int-5x\sin(2x)dx & = & \frac{5}{2}x\cos(2x)-\frac{5}{4}\sin(2x)\\\\\therefore\int_{0}^{\frac{\pi}{4}}-5x\sin(2x)dx & = & [\frac{5}{2}x\cos(2x)-\frac{5}{4}\sin(2x)]_{0}^{\frac{\pi}{4}}\\\end{eqnarray*}$

I'm sure you can take it from there. Lemme know if you have any questions about the working!

Thorium · « **Reply #5350 on:** July 13, 2014, 07:13:29 pm »

Quote from: jessss0407 on July 13, 2014, 06:53:07 pm

Hey guys! I have a question..

Find the derivative of y=5x cos(2x)
hence use your result to find the exact value of $\int_{0}^{pi/4} (-2xsin2x)dx$

Thanks!

I had a go, the solution is attached

Hope that helps:)

keltingmeith · « **Reply #5351 on:** July 13, 2014, 07:14:36 pm »

First thing's first - we need to derive it! We have a product of two functions, so let's apply the product rule:

$\frac{dy}{dx}=5x\cdot (-2sin(2x) + cos(2x)\cdot 5 = 5cos(2x) - 10xsin(2x)$

Now, the next part can only be solved with integration by parts - normally. See, methods has this little trick called "integration by recognition", which is baby integration by parts. Here's how you do these questions:

1) Re-write what you just found:
$\frac{d}{dx}5xcos(2x) = 5cos(2x) - 10xsin(2x)$

2) Set up an integral for the whole thing
$\int\limits_0^{\frac{\pi}{4}}\frac{d}{dx}5xcos(2x)\:dx=\int\limits_0^{\frac{\pi}{4}}5cos(2x)\:dx - \int\limits_0^{\frac{\pi}{4}}10xsin(2x)\:dx$

3) See if anything in there looks sort of like what you want to find, and solve for that:
$\int\limits_0^{\frac{\pi}{4}}\frac{d}{dx}5xcos(2x)\:dx-\int\limits_0^{\frac{\pi}{4}}5cos(2x)\:dx = -\int\limits_0^{\frac{\pi}{4}}10xsin(2x)\:dx<br />\\ 5\int\limits_0^{\frac{\pi}{4}}-2xsin(2x)\:dx=\int\limits_0^{\frac{\pi}{4}}\frac{d}{dx}5xcos(2x)\:dx-\int\limits_0^{\frac{\pi}{4}}5cos(2x)\:dx<br />\\ \int\limits_0^{\frac{\pi}{4}}-2xsin(2x)\:dx=\frac{1}{5}(\int\limits_0^{\frac{\pi}{4}}\frac{d}{dx}5xcos(2x)\:dx-\int\limits_0^{\frac{\pi}{4}}5cos(2x)\:dx)$

4) Integrate, knowing that the integral and derivative are basically the inverse of each other and just cancel out (but remember the terminals!)

EDIT: Beaten by Zealous, but I'll keep this here just because I've outlined the steps.

AngelWings · « **Reply #5352 on:** July 13, 2014, 09:14:22 pm »

Quote from: AngelWings on July 13, 2014, 12:32:09 pm

Please help me with this question. I know there's something I'm missing here that should be obvious.

Thanks in advance.

In case you wanted the rest of the question, but Q4d is the one I'm stuck on. Please help!

Thanks in advance.

knightrider · « **Reply #5353 on:** July 13, 2014, 11:17:38 pm »

How would you find the inverse of each of the following?

lzxnl · « **Reply #5354 on:** July 14, 2014, 12:37:46 am »

Quote from: knightrider on July 13, 2014, 11:17:38 pm

How would you find the inverse of each of the following?

Remember what the definitions of logs and exponentials are? I'll do the first one for you.

f(x) = 3^x
y=3^x
Swap x and y
x = 3^y

By definition of log, y = log₃ x

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Author Topic: VCE Methods Question Thread! (Read 5782416 times) Tweet Share

Conic

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