Author Topic: Mathematics Question Thread (Read 1307933 times) Tweet Share

mirakhiralla · « **Reply #3330 on:** March 12, 2018, 06:51:06 pm »

Heyyy!
Can someone help me out with this question
Thank you (:

RuiAce · « **Reply #3331 on:** March 12, 2018, 06:53:56 pm »

Quote from: mirakhiralla on March 12, 2018, 06:51:06 pm

Heyyy!
Can someone help me out with this question
Thank you (:

The attachment is kinda small and becomes blurry when zoomed in. What was the source of the question?

(So I can locate it myself, to save you from reuploading)

mirakhiralla · « **Reply #3332 on:** March 12, 2018, 07:02:37 pm »

Sorry,
I reuploaded the wrong question haahah

RuiAce · « **Reply #3333 on:** March 12, 2018, 07:08:21 pm »

$\text{The gradient of that line is }m = \frac{4-0}{0 - (-2/3)} = 6\\ \text{and its }y\text{-intercept is at }y=4.$
$\text{So using }y=mx+b,\text{ the equation of }f^{\prime\prime}(x)\\ \text{will just be }f^{\prime\prime}(x) = 6x + 4$
Note that we could've used $ y - y_1 = m(x-x_1) $ with either the point $ (0,4) $ or $ \left( -\frac23, 0 \right) $. But that was overkill for this question, since they gave us the $y$-intercept.
$\text{This means, that upon integrating, we have}\\ f^\prime(x) = 3x^2 + 4x + C_1.\\ \text{We're also told that at }x=2, f^\prime(x) = 1.\\ \therefore 1 = 12 + 8 + C_1 \implies C_1 = -19$
$\text{So }f^\prime(x) = 3x^2+4x-19$
From here, just do the same thing again. Integrate to get an expression for $ f(x) $, and to solve for the second constant $ C_2$ just use the fact that the curve $y=f(x)$ passes through (2,4)

Mate2425 · « **Reply #3334 on:** March 12, 2018, 11:36:46 pm »

Hi Could i please have some help with the following question:
Q) Find the exact area bounded by the curves y = sinx and y=cos x in the domain 0 < = X <= 2 pie. (Where 'arrows' are signs greater than/equal and less than/equal.

Thank you

Sine · « **Reply #3335 on:** March 13, 2018, 12:20:22 am »

Quote from: Mate2425 on March 12, 2018, 11:36:46 pm

Hi Could i please have some help with the following question:
Q) Find the exact area bounded by the curves y = sinx and y=cos x in the domain 0 < = X <= 2 pie. (Where 'arrows' are signs greater than/equal and less than/equal.

Thank you

$sin(x) = cos(x)$ $tan(x) = 1$ $x = \frac{\pi}{4}, \frac{5\pi}{4}$ $\text{between these x values we can see a bound region}$ $\text{however there still is some areas not accounted for}$ $\text{using symmetry properties the remaining area is the same}$ $\text{as what will be calculated for now so we just multiply by 2}$ $2\int_\frac{\pi}{4}^\frac{5\pi}{4} \! (sin(x) - cos(x)) \, \mathrm{d}x.$ $=2[-cos(x) - sin(x)]\Big|_\frac{\pi}{4}^\frac{5\pi}{4}$ $= 2((-cos(\frac{5\pi}{4}) -sin(\frac{5\pi}{4})) - (-cos(\frac{\pi}{4}) - sin(\frac{\pi}{4})))$ $= 2((- (-\frac{1}{\sqrt{2}}) - (-\frac{1}{\sqrt{2}})) - (-(\frac{1}{\sqrt{2}}) - (\frac{1}{\sqrt{2}})))$

$=4\sqrt{2} \text{squared units}$

Mate2425 · « **Reply #3336 on:** March 13, 2018, 07:31:11 am »

Quote from: Sine on March 13, 2018, 12:20:22 am

$sin(x) = cos(x)$ $tan(x) = 1$ $x = \frac{\pi}{4}, \frac{5\pi}{4}$ $\text{between these x values we can see a bound region}$ $\text{however there still is some areas not accounted for}$ $\text{using symmetry properties the remaining area is the same}$ $\text{as what will be calculated for now so we just multiply by 2}$ $2\int_\frac{\pi}{4}^\frac{5\pi}{4} \! (sin(x) - cos(x)) \, \mathrm{d}x.$ $=2[-cos(x) - sin(x)]\Big|_\frac{\pi}{4}^\frac{5\pi}{4}$ $= 2((-cos(\frac{5\pi}{4}) -sin(\frac{5\pi}{4})) - (-cos(\frac{\pi}{4}) - sin(\frac{\pi}{4})))$ $= 2((- (-\frac{1}{\sqrt{2}}) - (-\frac{1}{\sqrt{2}})) - (-(\frac{1}{\sqrt{2}}) - (\frac{1}{\sqrt{2}})))$ $=4\sqrt{2} \text{squared units}$

Thank you

gilliesb18 · « **Reply #3337 on:** March 21, 2018, 02:13:26 pm »

Hello,
Ok so I have a few questions re logs n exponentials...
1. what is the derivative of log_(10)x? and..
2. Find the equation of the tangent to the curve y= log_(e)x at the point (2, log_e(2))
I've been working on derivatives of logs etc. but these have just confused me a tad!!

Thanks heaps...

Opengangs · « **Reply #3338 on:** March 21, 2018, 03:20:16 pm »

Quote from: gilliesb18 on March 21, 2018, 02:13:26 pm

Hello,
Ok so I have a few questions re logs n exponentials...
1. what is the derivative of log_(10)x? and..
2. Find the equation of the tangent to the curve y= log_(e)x at the point (2, log_e(2))
I've been working on derivatives of logs etc. but these have just confused me a tad!!

Thanks heaps...

$\text{Recall the logarithmic property: } \log_{10}x = \frac{\ln(x)}{\ln(10)} \\ \text{Differentiating } \log_{10}x \text{, we get: } \\ \frac{d}{dx}\left(\log_{10}x\right) = \frac{d}{dx}\left(\frac{\ln(x)}{\ln(10)}\right) \\ \frac{d}{dx}\left(\log_{10}x\right) = \frac{1}{\ln(10)} \cdot \frac{1}{x} = \frac{1}{x\log(10)}$

Hopefully, this helps with the intuition behind logarithmic differentiation.
If you're still stuck on part (2), post all relevant working out and ask specifically where you're confused.

gilliesb18 · « **Reply #3339 on:** March 21, 2018, 03:54:09 pm »

Quote from: Opengangs on March 21, 2018, 03:20:16 pm

$\text{Recall the logarithmic property: } \log_{10}x = \frac{\ln(x)}{\ln(10)} \\ \text{Differentiating } \log_{10}x \text{, we get: } \\ \frac{d}{dx}\left(\log_{10}x\right) = \frac{d}{dx}\left(\frac{\ln(x)}{\ln(10)}\right) \\ \frac{d}{dx}\left(\log_{10}x\right) = \frac{1}{\ln(10)} \cdot \frac{1}{x} = \frac{1}{x\log(10)}$

Ohhh ok rightio thanks.... I just forgot that change of base thingo....

Thanks heaps!!

Mate2425 · « **Reply #3340 on:** March 23, 2018, 08:20:41 am »

Hi could please have some help with the following questions:

A) Evaluate (2x^3 -x^2 +5x +3) ÷ x with definite integral between 1 and 2

B) Use trapezoidal rule with 4 strips to find the area bounded by the curve y=In (x^2 -1), x-axis and the lines x =3 and x=5.

Thank you very much

RuiAce · « **Reply #3341 on:** March 23, 2018, 08:45:58 am »

Quote from: Mate2425 on March 23, 2018, 08:20:41 am

Hi could please have some help with the following questions:

A) Evaluate (2x^3 -x^2 +5x +3) ÷ x with definite integral between 1 and 2

B) Use trapezoidal rule with 4 strips to find the area bounded by the curve y=In (x^2 -1), x-axis and the lines x =3 and x=5.

Thank you very much

I'll just start you off.
$\int_1^2 \frac{2x^3-x^2+5x+3}\,dx = \int_1^2 2x^2-x+5 + \frac{3}{x}\, dx$
__________________
$\text{Using 4 subintervals, we have}\\ \int_3^5 f(x)\,dx \approx \frac{5-3}{2\times 4}[f(3) + f(5) + 2(f(3.5)+f(4)+f(4.5))]$
where $f(x) = \ln (x^2-1)$.

If you have further problems you should post up any progress you've made thus far.

LaraC · « **Reply #3342 on:** March 23, 2018, 11:37:01 am »

Hello

Could someone please help me to differentiate loge(2x+4)(3x-1)??
I know I should be using function of a function rule, but have completely confused myself!!

Thanks!

jazzycab · « **Reply #3343 on:** March 23, 2018, 12:26:20 pm »

Quote from: LaraC on March 23, 2018, 11:37:01 am

Hello

Could someone please help me to differentiate loge(2x+4)(3x-1)??
I know I should be using function of a function rule, but have completely confused myself!!

Thanks!

Is your question to differentiate $\log_e{\left(\left(2x+4\right)\left(3x-1\right)\right)}=\log_e{\left(6x^2+10x-4\right)}$ or $\log_e{\left(2x+4\right)}\left(3x-1\right)=\left(3x-1\right)\times\log_e{\left(2x+4\right)}$?
Given your comment at the end, assuming the former:
$\begin{align*}y&=\log_e{\left(6x^2+10x-4\right)}\\\text{Let }u&=6x^2+10x-4\\\therefore y&=\log_e{\left(u\right)}\\\frac{dy}{dx}&=\frac{dy}{du}\cdot\frac{du}{dx}\end{align*}$
From here, just evaluate each derivative, multiply them together, then substitute $u=6x^2+10x-4$ back in.

If it is the latter expression, you would need to use the product and chain rules.

LaraC · « **Reply #3344 on:** March 23, 2018, 12:30:29 pm »

Oh sorry....how stupid of me!

No it must be the product rule not the function of a function......the brackets are multiplied by each other and by loge.....they are not all in one big lot of brackets if that makes sense?

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