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Nithyashrri Saravanan
Subject: Maths
, asked on 19/3/18
Q. Find the dimension of a rectangle of perimeter 36 cm, which will sweep out a volume as large as possible , when revolved about one of its sides. Also, find the maximum value.
Answer
1
Archit
Subject: Maths
, asked on 19/3/18
Prove this as an increasing function
$\mathbf{9}\mathbf{.}Provethaty=\frac{4\mathrm{sin}\theta}{(2+\mathrm{cos}\theta )}-\theta isanincrea\mathrm{sin}gfunctionof\theta in\left[0,\frac{\mathrm{\pi}}{2}\right].$
Answer
1
Helen Elma
Subject: Maths
, asked on 18/3/18
Show that
f
(
x
) = sin
x
(1 + cos
x
) is maximum when
$a=\frac{\pi}{3}$
in the interval [0, π].
Answer
1
Arnapurna Paikaray
Subject: Maths
, asked on 18/3/18
Please explain me Q12 (c) of Exercise 6.2 .
Answer
3
Arnapurna Paikaray
Subject: Maths
, asked on 18/3/18
Please explain (c ) again.Why we took pie instead of 0?
Answer
1
Vibhav
Subject: Maths
, asked on 18/3/18
In this answer the point (4,-8/3) does not satisfy the equation 6y = x^2
So how is this one of the solutions???
Answer
2
Vibhav
Subject: Maths
, asked on 18/3/18
The points on the curve 9
y
^{2}
=
x
^{3}
, where the normal to the curve makes
equal
intercepts with the axes are
(A)
$\left(4,\pm \frac{8}{3}\right)$
(B)
$4,\frac{-8}{3}$
(C)
$\left(4,\pm \frac{3}{8}\right)$
(D)
$\left(\pm 4,\frac{3}{8}\right)$
Answer
1
Suroj Dey
Subject: Maths
, asked on 18/3/18
If f'(x) has no roots, does that mean f(x) will always have at least one root?
Answer
1
Varnika Dhiman
Subject: Maths
, asked on 17/3/18
In the solution of the following question , I want to know how the marked (in blue bracket) step came ?
Q. Find the intervals in which
$f\left(x\right)=\mathrm{sin}3x-\mathrm{cos}3x,0x\pi $
is strictly increasing or strictly decreasing?
Answer
1
Varnika Dhiman
Subject: Maths
, asked on 17/3/18
How did we get to know it's "strictly" decreasing or "strictly" increasing ?!..
Why not just decreasing/increasing ?!..
Answer
1
Arnapurna Paikaray
Subject: Maths
, asked on 17/3/18
How area is calculated in this question? Please explain .Dont send links .
Answer
1
Arnapurna Paikaray
Subject: Maths
, asked on 17/3/18
Please explain how it is done. Don't send links .
Solutions:
Ellipse, the area of triangle ABC (A) is given by.
$A=\frac{1}{2}\overline{)a\left(\frac{2b}{a}\sqrt{{a}^{2}-{x}_{1}^{2}}\right)+\left(-{x}_{1}\right)}\left(-\frac{b}{a}\sqrt{{a}^{2}-{x}_{1}^{2}}\right)+\left(-{x}_{1}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow A=b\sqrt{{a}^{2}-{x}_{1}^{2}}+{x}_{1}\frac{b}{a}\sqrt{{a}^{2}-{x}_{1}^{2}}\phantom{\rule{0ex}{0ex}}\therefore \frac{dA}{d{x}_{1}}=\frac{-2{x}_{1}b}{2\sqrt{{a}^{2}-{x}_{1}^{2}}}+\frac{\mathbf{b}}{\mathbf{a}}\sqrt{{a}^{2}-{x}_{1}^{2}}-\frac{2b{x}_{1}^{2}}{a2\sqrt{{a}^{2}-{x}_{1}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{b}{a\sqrt{{a}^{2}-{x}_{1}^{2}}}\left[-{x}_{1}a+\left({a}^{2}-{x}_{1}^{2}\right)-{x}_{1}^{2}\right]\phantom{\rule{0ex}{0ex}}Now.\frac{dA}{d{x}_{1}}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow -2{x}_{1}^{2}-{x}_{1}a+{a}^{2}=+\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{1}=\frac{a\pm \sqrt{{a}^{2}-4\left(-1\right)\left({a}^{2}\right)}}{2\left(-2\right)}$
Answer
1
Arnapurna Paikaray
Subject: Maths
, asked on 17/3/18
Please explain how perpendicular distance came . Dont provide link please.
Solution:
The equation of the normal at a given point (x,y) is given by,
$v-a\mathrm{sin}\theta +a\theta \mathrm{cos}\theta =\frac{-1}{ten\theta}\left(x-a\mathrm{cos}\theta -a\theta \mathrm{sin}\theta \right)\phantom{\rule{0ex}{0ex}}\Rightarrow y\mathrm{sin}\theta -a{\mathrm{sin}}^{2}\theta +a\theta \mathrm{sin}\theta \mathrm{cos}\theta =x\mathrm{cos}\theta +a{\mathrm{cos}}^{2}\theta \phantom{\rule{0ex}{0ex}}\Rightarrow x\mathrm{cos}\theta +y\mathrm{sin}g\theta -a\left({\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \right)=\phantom{\rule{0ex}{0ex}}\Rightarrow x\mathrm{cos}\theta +y\mathrm{sin}\theta -a=0$
Now, the perpendicular distance of he normal from the origin is.
$\frac{\left|-a\right|}{\sqrt{{\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta}}=\frac{\left|-a\right|}{\sqrt{1}}=\left|a\right|,\mathrm{which}\mathrm{is}\mathrm{independent}\mathrm{of}$
Hence, the perpendicular distance of the normal from the origin is constant.
Answer
1
Megha Prasadan
Subject: Maths
, asked on 16/3/18
Q) Why didn't we change the units or keep it same?
Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x
m away from the wall.
Then, by Pythagoras theorem, we have:
x
^{2}
+ y
^{2}
= 25 [Length of the ladder = 5 m]
$y=\sqrt{25-{x}^{2}}$
Then, the rate of change of height (y) with respect to time (t) is given by,
$\frac{dy}{dx}=\frac{-x}{\sqrt{25-{x}^{2}}}.\frac{dx}{dt}$
It is given that dx/dt = 2 cm/s
$\frac{dy}{dt}=\frac{-2x}{\sqrt{25-{x}^{2}}}$
Now, when x = 4 m, we have:
$\frac{dy}{dt}=\frac{-2\times 4}{\sqrt{25-16}}=-\frac{8}{3}$
Hence, the height of the ladder on the wall is decreasing at the rate of 8/3 cm/s.
Answer
5
Arnapurna Paikaray
Subject: Maths
, asked on 16/3/18
Please dont provide link.
Answer
3
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Q. Find the dimension of a rectangle of perimeter 36 cm, which will sweep out a volume as large as possible , when revolved about one of its sides. Also, find the maximum value.

$\mathbf{9}\mathbf{.}Provethaty=\frac{4\mathrm{sin}\theta}{(2+\mathrm{cos}\theta )}-\theta isanincrea\mathrm{sin}gfunctionof\theta in\left[0,\frac{\mathrm{\pi}}{2}\right].$

f(x) = sinx(1 + cosx) is maximum when $a=\frac{\pi}{3}$ in the interval [0, π].So how is this one of the solutions???

y^{2}=x^{3}, where the normal to the curve makes equal intercepts with the axes are(A) $\left(4,\pm \frac{8}{3}\right)$ (B) $4,\frac{-8}{3}$ (C) $\left(4,\pm \frac{3}{8}\right)$ (D) $\left(\pm 4,\frac{3}{8}\right)$

In the solution of the following question , I want to know how the marked (in blue bracket) step came ?Q. Find the intervals in which $f\left(x\right)=\mathrm{sin}3x-\mathrm{cos}3x,0x\pi $ is strictly increasing or strictly decreasing?

How did we get to know it's "strictly" decreasing or "strictly" increasing ?!..

Why not just decreasing/increasing ?!..

Solutions:

Ellipse, the area of triangle ABC (A) is given by.

$A=\frac{1}{2}\overline{)a\left(\frac{2b}{a}\sqrt{{a}^{2}-{x}_{1}^{2}}\right)+\left(-{x}_{1}\right)}\left(-\frac{b}{a}\sqrt{{a}^{2}-{x}_{1}^{2}}\right)+\left(-{x}_{1}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow A=b\sqrt{{a}^{2}-{x}_{1}^{2}}+{x}_{1}\frac{b}{a}\sqrt{{a}^{2}-{x}_{1}^{2}}\phantom{\rule{0ex}{0ex}}\therefore \frac{dA}{d{x}_{1}}=\frac{-2{x}_{1}b}{2\sqrt{{a}^{2}-{x}_{1}^{2}}}+\frac{\mathbf{b}}{\mathbf{a}}\sqrt{{a}^{2}-{x}_{1}^{2}}-\frac{2b{x}_{1}^{2}}{a2\sqrt{{a}^{2}-{x}_{1}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{b}{a\sqrt{{a}^{2}-{x}_{1}^{2}}}\left[-{x}_{1}a+\left({a}^{2}-{x}_{1}^{2}\right)-{x}_{1}^{2}\right]\phantom{\rule{0ex}{0ex}}Now.\frac{dA}{d{x}_{1}}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow -2{x}_{1}^{2}-{x}_{1}a+{a}^{2}=+\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{1}=\frac{a\pm \sqrt{{a}^{2}-4\left(-1\right)\left({a}^{2}\right)}}{2\left(-2\right)}$

Solution:

The equation of the normal at a given point (x,y) is given by,

$v-a\mathrm{sin}\theta +a\theta \mathrm{cos}\theta =\frac{-1}{ten\theta}\left(x-a\mathrm{cos}\theta -a\theta \mathrm{sin}\theta \right)\phantom{\rule{0ex}{0ex}}\Rightarrow y\mathrm{sin}\theta -a{\mathrm{sin}}^{2}\theta +a\theta \mathrm{sin}\theta \mathrm{cos}\theta =x\mathrm{cos}\theta +a{\mathrm{cos}}^{2}\theta \phantom{\rule{0ex}{0ex}}\Rightarrow x\mathrm{cos}\theta +y\mathrm{sin}g\theta -a\left({\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \right)=\phantom{\rule{0ex}{0ex}}\Rightarrow x\mathrm{cos}\theta +y\mathrm{sin}\theta -a=0$

Now, the perpendicular distance of he normal from the origin is.

$\frac{\left|-a\right|}{\sqrt{{\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta}}=\frac{\left|-a\right|}{\sqrt{1}}=\left|a\right|,\mathrm{which}\mathrm{is}\mathrm{independent}\mathrm{of}$

Hence, the perpendicular distance of the normal from the origin is constant.

Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x

m away from the wall.

Then, by Pythagoras theorem, we have:

x

^{2}+ y^{2}= 25 [Length of the ladder = 5 m]$y=\sqrt{25-{x}^{2}}$

Then, the rate of change of height (y) with respect to time (t) is given by,

$\frac{dy}{dx}=\frac{-x}{\sqrt{25-{x}^{2}}}.\frac{dx}{dt}$

It is given that dx/dt = 2 cm/s

$\frac{dy}{dt}=\frac{-2x}{\sqrt{25-{x}^{2}}}$

Now, when x = 4 m, we have:

$\frac{dy}{dt}=\frac{-2\times 4}{\sqrt{25-16}}=-\frac{8}{3}$

Hence, the height of the ladder on the wall is decreasing at the rate of 8/3 cm/s.