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What will be the average rate of change of the function [y = 16 – x^{2}] between x = 3 and x = 4?
Let, y = f(x) = 16 – x^{2}
If x changes from 3 to 4, then, δx = 4 – 3 = 1
Again f(4) = 16 – 4^{2} = 0
And f(3) = 16 – 3^{2} = 7
Therefore, δy = f(4) – f(3) = 0 – 7 = 7
Hence, the average rate of change of the function between x = 3 and x = 4 is:
δy/δx = 7/1 = 7.
What will be the approximate change in the surface area of a cube of side xm caused by increasing the side by 2%.
Let the edge of the cube be x. Given that dx or Δx is equal to 0.02x(2%).
The surface area of the cube is A=6x^{2}
Differentiating w.r.t x, we get
dA/dx =12x
dA= (dA/dx)Δx=12x(0.02x)=0.24x^{2}
Hence, the approximate change in volume is 0.24x^{2}.
Let −cot^{−1}x=t
Differentiating w.r.t x, we get
=e^{t}
Replacing t with cot^{1}x, we get
To find
We know that
By simplifying it we get,
Now equating the coefficients we get A = 0, B = 0, C=1.
Therefore after integrating we get logx– (1/2)log(x^{2}+1) + C.
By simplifying, it we get
By solving the equations, we get, A+B=0 and 3A3B=1
By solving these 2 equations, we get values of A=1/6 and B=1/6.
Now by putting values in the equation and integrating it we get value,
What will be the average rate of change of the function [y = 16 – x^{2}] at x = 4?
Let, y = f(x) = 16 – x^{2}
dy/dx = 2x
Now, [dy/dx]_{x = 4} = [2x]_{x = 4}
So, [dy/dx]_{x = 4} = 8
Find the approximate value of f(4.04), where f(x)=7x^{3}+6x^{2}4x+3.
Let x=4 and Δx=0.04
Then, f(x+Δx)=7(x+Δx)^{3}+7(x+Δx)^{2}4(x+Δx)+3
Δy=f(x+Δx)f(x)
∴f(x+Δx)=Δy+f(x)
Δy=f'(x)Δx
⇒ f(x+Δx)=f(x)+f’ (x)Δx
Here, f'(x)=21x^{2}+12x4
f(4.04)=(7(4)^{3}+6(4)^{2}4(4)+3)+(21(4)^{2}+12(4)4)(0.04)
f(4.04)=(7(4)3+6(4)24(4)+3)+(21(4)2+12(4)4)(0.04)
f(4.04)=531+380(0.04)=546.2
Let x^{5}+9=t
Differentiating w.r.t x, we get
5x^{4} dx=dt
Replacing t with x^{5}+9, we get
To find
What will be the value of the coordinate whose position of a particle moving along the parabola y^{2} = 4x at which the rate at of increase of the abscissa is twice the rate of increase of the ordinate?
Here, y^{2} = 4x ….(1)
Let, (x, y) be the position of the particle moving along the parabola (1) at time t.
Now, differentiating both sides of (1) with respect to t, we get:
2y(dy/dt) = 4(dx/dt)
Or, y(dy/dt) = 2(dy/dt) ……….(2)
By question, dx/dt = 2 * dy/dt ……….(3)
From (2) and (3) we get, y(dy/dt) = 2 * 2 dy/dt
Or, y = 4
Putting y = 4 in (1) we get, 4^{2} = 4x
So, x = 4
Thus, the coordinate of the particle is (4, 4).
Let y=(x)^{1/3}. Let x=125 and Δx=2
Then, Δy=(x+Δx)^{1/3}x^{1/3}
Δy=(127)^{1/3}(125)^{1/3}
(127)^{1/3}=Δy+5
dy is approximately equal to Δy is equal to
dy=2/75=0.0267
∴ The approximate value of (127)^{1/3} is 5+0.0267=5.0267
Let √x = t
Differentiating w.r.t x,we get
=12(cost)=12 cost
Replacing t with √x, we get
To find ∫(2+x) x √x dx
It is a form of the given partial fraction which can also be written as and is further used to solve integration by partial fractions numerical.
The time rate of change of the radius of a sphere is 1/2π. When it’s radius is 5cm, what will be the rate of change of the surface of the sphere with time?
Let, r be the radius and s be the area of the surface of the sphere at time t.
By question, dr/dt = 1/2π
Now, s = 4πr^{2};
Thus, ds/dt = 4π * 2r(dr/dt) = 8πr(dr/dt)
When r = 5cm and dr/dt = 1/2π
Then ds/dt = 8π*5*(1/2π) = 20
Thus, correct answer is 20 sq cm.
Find the approximate change in the volume of cube of side xm caused by increasing the side by 6%.
We know that the volume V of a cube is given by V=x^{3}
Differentiating w.r.t x, we get
dV/dx = 3x^{2}
dV = (dV/dx)Δx=3x^{2} Δx
dV=3x^{2} (6x/100)=0.18x^{3}
Therefore, the approximate change in volume is 0.18x^{3}.
Let 1+x^{4}=t
4x^{3} dx=dt
=5 logt
Replacing t with 1+x^{4}, we get
= 5log(1+x^{4})+C
To find:
Now equating, (x^{2}+x+1) = A (x^{2}+1) + (Bx+C) (x+2)
After equating and solving for coefficient we get values, now putting these values in the equation we get,
Hence it comes,
A solid cube changes its volume such that its shape remains unchanged. For such a cube of unit volume, what will be the value of rate of change of volume?
Let x be the length of a side of the cube.
If v be the volume and s the area of any face of the cube, then
v = x^{3} and s = x^{2}
Thus, dv/dt = dx^{3}/dt = 3x^{2 }(dx/dt)
And ds/dt = dx^{2}/dt = 2x(dx/dt)
Now, (dv/dt)/(ds/dt) = 3x/2
Or, dv/dt = (3x/2)(ds/dt)
Now, for a cube of unit volume we have,
v = 1
⇒ x = 1 [as, x is real]
Therefore, for a cube of unit volume [i.e. for x = 1], we get,
dv/dt = (3/2)(ds/dt)
Thus the rate of change of volume = 3/2*(rate of change of area of any face of the cube)
Let y=x^{1/4}. Let x=81 and Δx=1
Then, Δy=(x+Δx)^{1/4}x^{1/4}
Δy=82^{1/4}81^{1/4}
82^{1/4}=Δy+3
dy is approximately equal to Δy is equal to
dy = (dy/dx)Δx
∴ The approximate value of 82^{1/4} is 3+0.00925=3.00925
Let x^{3}=t
3x^{2} dx=dt
x^{2} dx=dt/3
Replacing t with x^{3}, we get
To find ∫sin2x+e^{3x}−cos3xdx
∫sin2x+e^{3x}−cos3xdx=∫sin2xdx+∫e^{3x}dx−∫cos3xdx
∫sin2x+e^{3x}−cos3xdx=
As it represents the identity (b+a)^{2} it satisfies the identity (b+a)^{2} = (a^{2 }+ b^{2} +2ab) and is not linear, cubic or an imaginary equation so the correct option is Identity Equation.
A 5 ft long man walks away from the foot of a 12(½) ft high lamp post at the rate of 3 mph. What will be the rate at which the shadow increases?
Let, AB be the lamppost whose foot is A, and B is the source of light, and given (AB)’ = 12(½) ft.
Let MN denote the position of the man at time t where (MN)’ = 5ft.
Join BN and produce it to meet AM(produced) at P.
Then the length of man’s shadow= (MP)’
Assume, (AM)’ = x and (MP)’ = y. Then, (PA)’ = (AM)’ + (MP)’ = x + y
And dx/dt = velocity of the man = 3
Clearly, triangles APB and MPN are similar.
Thus, (PM)’/(MN)’ = (PA)’/(AB)’
Or, y/5 = (x + y)/12(½)
Or, (25/2)y = 5x + 5y
Or, 3y = 2x
Or, y = (2/3)x
Thus, dy/dt = (2/3)(dx/dt)
As, dx/dt = 2,
= 2/3*3 = 2mph
Find the approximate error in the volume of the sphere if the radius of the sphere is measured to be 6cm with an error of 0.07cm.
Let x be the radius of the sphere.
Then, x=6cm and Δx=0.07cm
The volume of a sphere is given by V= (4/3)πx^{3}
dV=4×π×6^{2}×0.07
dV=10.08π cm^{3}
Let cos^{1}x=t
Differentiating w.r.t x, we get
= t^{2}/2
Replacing t with cos^{1}x,we get
To find (ax^{2}+b)^{2}
∫(ax^{2}+b)^{2}dx=∫(a^{2}x^{4}+b^{2}+2ax^{2}b)dx
∫(ax^{2}+b)^{2}dx=∫a^{2}x^{4}dx+∫b^{2}dx+2∫ax^{2}bdx
∫(ax^{2}+b)^{2}dx=a^{2}∫x^{4}dx+b^{2}∫dx+2ab∫x^{2}dx
For the given equation (x+2) (x+4) = x^{2} + 6x + 8, how many values of x satisfies this equation?
If we solve the L.H.S. (Left Hand Side) of the equation, we get the following value.
(x+2) (x+4) = x^{2} + 4x + 2x + 8 = x^{2 }+ 6x + 8.
This value is same as the R.H.S. (Right Hand Side).
So, all the values of x satisfy the equality.
A ladder 20 ft long leans against a vertical wall. If the top end slides downwards at the rate of 2ft per second, what will be the rate at which the slope of the ladder changes?
Let the height on the wall be x and laser touches the ground at distance y from the wall. The length of the ladder is 20ft.
By Pythagoras theorem:
x^{2} + y^{2} = 400
Differentiating with respect to t:
2x(dx/dt) + 2y(dy/dt) = 0
Dividing throughout by 2:
x (dx/dt) + y (dy/dt) = 0
Now, dx/dt = 2 ft /s. negative because downwards
x(2) + y (dy/dt) = 0 ………..(1)
When lower end is 12 ft from wall, let us find x:
x^{2} + 12^{2} = 400
x^{2} = 400 – 144= 256
x = 16
x(2) + y (dy/dt) = 0 from (1)
16(2) + 12 (dy/dt) = 0
32 + 12(dy/dt) = 0
dy/dt = (32/12) = (8/3)
Thus, lower end moves on a horizontal floor when it is 12 ft from the wall at the rate of 8/3 ft/s
Now, assume that the ladder makes an angle θ with the horizontal plane at time t.
If, m be the slope of the ladder at time t, then,
m = tanθ = x/y
Thus, dm/dt = d/dt(x/y) = [y(dx/dt) – x(dx/dt)]/y^{2}
Therefore, the rate of change of slope of ladder is,
[dm/dt]_{y = 2} = [12*(2) – 16*(8/3)]/(12)^{2}
Now, putting the value of x = 16, when y = 12 and dx/dt = 2, dy/dt = 8/3
We get, [dm/dt]_{y = 12} = [12(2) – 16(8/3)]/(12)^{2} = 25/54
A particle moving in a straight line covers a distance of x cm in t second, where x = t^{3} + 6t^{2} – 15t + 18. What will be the acceleration of the particle at the end of 2 seconds?
We have, x = t^{3} + 6t^{2} – 15t + 18
Let, ‘v’ be the velocity of the particle and ‘a’ be the acceleration of the particle at the end of t seconds. Then,
v = dx/dt = d/dt(t^{3} + 6t^{2} – 15t + 18)
So, v = 3t^{2} + 12t – 15
Therefore, a = dv/dt = d/dt(3t^{2} + 12t – 15)
So, a = 6t + 12
Thus, acceleration of the particle at the end of 2 seconds is, [dv/dt]_{t = 2} = 6(2) + 12 = 24cm/sec^{2}.
Water is flowing into a right circular conical vessel, 45 cm deep and 27 cm in diameter at the rate of 11 cc per minute. How fast is the water level rising when the water is 30 cm deep?
Let ‘r’ be the radius and ‘h’ be the height of the water level at time t.
Then the volume of water level ‘V’ at time t is given by,
V = 1/3 (πr^{2}h) ……….(1)
Given the radius of the base of the cone is = (OA)’ = 27/2 cm and its height = (OB)’ = 45cm.
Again at time t, the radius of the water level = r = (CD)’ and its height = h = (CB)’
Clearly, the triangle OAB and CBD are similar.
Therefore, (CD)’/(CB)’ = (OA)’/(OB)’
Or, r/h = (22/7)/45 = 3/10
or, r = 3h/10
Thus, from (1) we get,
V = 1/3 π (3h/10)^{2} h = (3π/100)h^{3}
Thus, dV/dt = (3π/100)*3h^{2}(dh/dt)
= (9πh^{2}/100)(dh/dt)
When, h = 30cm, then, 11 = (9π/100) (30)^{2} (dh/dt) [as for all the values of t we have, dV/dt = 11]
Or, dh/dt = (11/(9π*9)) = 0.043
Thus, the rising rate of rising is 0.043 cm/minute.
A particle moving in a straight line covers a distance of x cm in t second, where x = t^{3} + 6t^{2} – 15t + 18. When does the particle stop?
We have, x = t^{3} + 6t^{2} – 15t + 18
The particle stops when dx/dt = 0
And, dx/dt = d/dt(t^{3} + 6t^{2} – 15t + 18)
3t^{2} + 12t – 15 = 0
⇒ t^{2} + 4t –5 = 0
⇒ (t – 1)(t + 5)= 0
Thus, t = 1 or t = 5
Hence, the particle stops at the end of 1 second.
A particle moving in a straight line covers a distance of x cm in t second, where x = t^{3} + 6t^{2} – 15t + 18. What will be the velocity of the particle at the end of 2 seconds?
We have, x = t^{3} + 6t^{2} – 15t + 18
Let, v be the velocity of the particle at the end of t seconds. Then, v = dx/dt = d/dt(t^{3} + 6t^{2} – 15t + 18)
So, v = 3t^{2} + 12t – 15
Thus, velocity of the particle at the end of 2 seconds is, [dx/dt]_{t = 2} = 3(2)^{2} + 12(2) – 15 = 21cm/sec.
If L_{1} and L_{2} have the direction ratios a_{1},b_{1},c_{1 }and a_{2},b_{2},c_{2} respectively then what is the angle between the lines?
If L_{1} and L_{2} have the direction ratios
a_{1},b_{1},c_{1 }and a_{2},b_{2},c_{2} respectively then the angle between the lines is given by
Find the value of p such that the lines
are at right angles to each other.
The angle between two lines is given by the equation
0 = p+1
p = 1
Find the value of p such that the lines are at right angles to each other.
We know that, if two lines are perpendicular to each other then, a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0
i.e.4(p)+(2)2+4(12)=0
4p448=0
4p=52
p = 52/4 = 13.
If the equations of two lines L_{1} and L_{2} are and then which of the following is the correct formula for the angle between the two lines?
Given that the equations of the lines are
∴ the angle between the two lines is given by
If two lines L_{1} and L_{2} are having direction cosines l_{1},m_{1},n_{1}and l_{2},m_{2},n_{2} respectively, then what is the angle between the two lines?
If two lines L_{1} and L_{2} are having direction cosines l_{1},m_{1},n_{1 }and l_{2},m_{2},n_{2} respectively, then the angle between the lines is given by cosθ=l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}
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