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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
9
.
P
r
o
v
e
t
h
a
t
y
=
4
sin
θ
(
2
+
cos
θ
)
-
θ
i
s
a
n
i
n
c
r
e
a
sin
g
f
u
n
c
t
i
o
n
o
f
θ
i
n
0
,
π
2
.
Answer
1
Helen Elma
Subject: Maths
, asked on 18/3/18
Show that
f
(
x
) = sin
x
(1 + cos
x
) is maximum when
a
=
π
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)
4
,
±
8
3
(B)
4
,
-
8
3
(C)
4
,
±
3
8
(D)
±
4
,
3
8
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
x
=
sin
3
x
-
cos
3
x
,
0
<
x
<
π
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
=
1
2
a
2
b
a
a
2
-
x
1
2
+
-
x
1
-
b
a
a
2
-
x
1
2
+
-
x
1
⇒
A
=
b
a
2
-
x
1
2
+
x
1
b
a
a
2
-
x
1
2
∴
d
A
d
x
1
=
-
2
x
1
b
2
a
2
-
x
1
2
+
b
a
a
2
-
x
1
2
-
2
b
x
1
2
a
2
a
2
-
x
1
2
=
b
a
a
2
-
x
1
2
-
x
1
a
+
a
2
-
x
1
2
-
x
1
2
N
o
w
.
d
A
d
x
1
=
0
⇒
-
2
x
1
2
-
x
1
a
+
a
2
=
+
⇒
x
1
=
a
±
a
2
-
4
-
1
a
2
2
-
2
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
sin
θ
+
a
θ
cos
θ
=
-
1
t
e
n
θ
x
-
a
cos
θ
-
a
θ
sin
θ
⇒
y
sin
θ
-
a
sin
2
θ
+
a
θ
sin
θ
cos
θ
=
x
cos
θ
+
a
cos
2
θ
⇒
x
cos
θ
+
y
sin
g
θ
-
a
sin
2
θ
+
cos
2
θ
=
⇒
x
cos
θ
+
y
sin
θ
-
a
=
0
Now, the perpendicular distance of he normal from the origin is.
-
a
cos
2
θ
+
sin
2
θ
=
-
a
1
=
a
,
which
is
independent
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
=
25
-
x
2
Then, the rate of change of height (y) with respect to time (t) is given by,
d
y
d
x
=
-
x
25
-
x
2
.
d
x
d
t
It is given that dx/dt = 2 cm/s
d
y
d
t
=
-
2
x
25
-
x
2
Now, when x = 4 m, we have:
d
y
d
t
=
-
2
×
4
25
-
16
=
-
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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What are you looking for?
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.
So how is this one of the solutions???
(A) (B) (C) (D)
Q. Find the intervals in which is strictly increasing or strictly decreasing?
Why not just decreasing/increasing ?!..
Solutions:
Ellipse, the area of triangle ABC (A) is given by.
Solution:
The equation of the normal at a given point (x,y) is given by,
Now, the perpendicular distance of he normal from the origin is.
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:
x2 + y2 = 25 [Length of the ladder = 5 m]
Then, the rate of change of height (y) with respect to time (t) is given by,
It is given that dx/dt = 2 cm/s
Now, when x = 4 m, we have:
Hence, the height of the ladder on the wall is decreasing at the rate of 8/3 cm/s.