Author Topic: Cobby's Methods Questions (Read 40702 times) Tweet Share

kamil9876 · « **Reply #120 on:** May 20, 2009, 11:54:01 am »

Nice

I believe I've seen an analysis question in specialist that uses that dot product approach.
You can also prove that the product of gradients of two perpendicular lines equals -1 using the dot product, so i guess that's where the two approaches are somewhat logically related

cobby · « **Reply #121 on:** May 22, 2009, 07:29:53 pm »

Hey guys,

can someone help me with this question.. It's preparation for the SAT's coming up.

The length of rectangle S is 20 percent longer than the length of rectangle R, and the width of rectangle S is 20 percent shorter than the width of rectangle R. The area of rectangle S is:

A) 20% greater than the area of rectangle R
B) 4% greater than the area of rectangle R
C) equal to the area of rectangle R
D) 4% less than the area of rectangle R
E) 20% less than the area of rectangle R

Thanks

pHysiX · « **Reply #122 on:** May 22, 2009, 08:56:57 pm »

D

Let length of rectangle R be "L" and width of rectangle R be "W"
Let length of rectangle S be "A" and width of rectangle S be "B"

From given information:
Length of rectangle S is 20 percent longer, which implies length of S = 1.2 length of W

i.e. A=1.2L

Width of rectangle S is 20 percent shorter, which implies width of S = 0.8 width of W

i.e. B=0.8W

Area of S = AB = 0.96LW
Area of R = LW

Difference in Areas = 0.04

Percentage = .04/1 (difference divided by original) x100

so therefore, S is 4% less

kamil9876 · « **Reply #123 on:** May 22, 2009, 09:03:33 pm »

"The SAT Reasoning Test (formerly Scholastic Aptitude Test and Scholastic Assessment Test) is a standardized test for college admissions in the United States"

whoa, plans to go to US.

anyways back to the question...

so lets denote our variables: $w_R, L_R,w_S,L_S$

$L_S=1.2L_R$
$W_S=0.8W_R$

multiply the equations together to get:

$L_S W_S=1.2*0.8 L_R W_R$
$A_S=0.96A_R$

So Area of S is 96% of the area of R. Hence it is 4% less.

cobby · « **Reply #124 on:** May 22, 2009, 09:12:46 pm »

Wow thanks guys!!!

And yes Kamil i plan to go to the states next year

kurrymuncher · « **Reply #125 on:** May 24, 2009, 07:37:29 pm »

Quote from: cobby on May 22, 2009, 09:12:46 pm

Wow thanks guys!!!

And yes Kamil i plan to go to the states next year

To cleeveland community college ahaha

cobby · « **Reply #126 on:** May 28, 2009, 10:06:57 pm »

Hey guys another one please

The surface area of a cube is changing at the rate of $8 cm^2/s$ . How fast is the volume changing when the surfrace area is $60 cm$ ?

Thanks guys

dekoyl · « **Reply #127 on:** May 28, 2009, 10:26:53 pm »

$S = 4\pi{r^2}$
$\frac{dS}{dt}=8 cm^2/s$
We want:
$\frac{dV}{dt}=\frac{dV}{dS}.\frac{dS}{dt}$
$V = \frac{4}{3}\pi{r^3}$
$\implies V = \frac{S}{3}\sqrt{\frac{S}{4\pi}$
$\implies V = \sqrt{\frac{S^3}{12\pi}$
$\frac{dV}{dS}=\frac{3S^2}{24\pi\sqrt{\frac{S^3}{12\pi}}}$
$\implies \frac{dV}{dt}=\frac{S^2}{\pi\sqrt{\frac{S^3}{12\pi}}}$

Now sub $S=60$ into the above.

Sorry. I might have bullshitted my way through because it's WTF-esque.

LOL ITS FOR A CUBE. I READ SPHERE. SORRY.

Flaming_Arrow · « **Reply #128 on:** May 28, 2009, 10:32:27 pm »

i got 2 rt[10] but i accidently exited my window after typing everything up, cbf now

Mao · « **Reply #129 on:** May 28, 2009, 11:03:31 pm »

$S=6r^2\implies r = \left(\frac{S}{6}\right)^{1/2}$

$V=r^3 = \left(\frac{S}{6}\right)^{3/2}$

$\frac{dV}{dt} = \frac{dS}{dt}\cdot \frac{dV}{dS}$

the rest is trivial

cobby · « **Reply #130 on:** May 31, 2009, 07:36:55 pm »

Hey guys, can someone please help me with this?

Find the following
$\frac{d(e^x f(x))}{dx}$

I don't even know where to start with this??

Thanks guys

d0minicz · « **Reply #131 on:** May 31, 2009, 07:42:55 pm »

do it like normal using the product rule
so theyre asking you to find the derivative of $e^x f(x)$
remember the derivative of f(x) is f'(x) and use that

Damo17 · « **Reply #132 on:** May 31, 2009, 07:43:38 pm »

Quote from: cobby on May 31, 2009, 07:36:55 pm

Hey guys, can someone please help me with this?

Find the following
$\frac{d(e^x f(x))}{dx}$

I don't even know where to start with this??

Thanks guys

use product rule:
$<br />\frac{d}{dx}(e^x f(x))= e^x f(x)+e^x f '(x)$
$= e^x(f(x)+ f '(x))$

cobby · « **Reply #133 on:** June 17, 2009, 06:06:35 pm »

Hey guys

Given that $y= e^{2t}$ and $x = e^t +1$

Find $\frac {dy}{dx}$ when $t=0$

Thanks

Mao · « **Reply #134 on:** June 17, 2009, 06:12:16 pm »

$\frac{dy}{dt} = 2e^{2t}$ , $\frac{dx}{dt} = e^t$

$\implies \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = 2e^t$

when t=0, $\frac{dy}{dx} = 2$

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