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Answered on 27/01/2022 Learn Applications of Derivatives

Pugazhendhi V

IT Technical Executive with 4 years of experience

Solution: Let length of rectangle =x cm and width of rectangle =y cm Given the length is decreasing at the rate of 3 cm/ minutes i.e. is decreasing w.r.t time dtdx=−3 cm/min.......(1) as x is decreasing and the width y is increasing at the rate of 2 cm/min i.e. y is increasing w.r.t... read more

Solution:

Let length of rectangle =x cm
and width of rectangle =y cm

Given the length is decreasing at the rate of 3 cm/ minutes i.e. is decreasing w.r.t time

dtdx=−3 cm/min.......(1) as x is decreasing and the width y is increasing at the rate of 2 cm/min
i.e. y is increasing w.r.t time
dy/dt=2 cm/min.........(2)

(1) Let P be the perimeter of rectangle
=2(l+width)
P=2(x+y)
dp/dt=2,(dx/dt+dy/dt)
Given : dx/dt=−3,dy/dt=2
∴dp/dt=2(−3+2)=2×−1=−2 cm/min
∴ perimeter is decreasing at the rate of 2 cm/min

(2) Let A be the area of the rectangle
A=l×width=xy⇒dA/dt=xdy/dt+ydtdx
⇒dA/dt=−3y+2x from (1) & (2)
dA/dt∣∣∣∣x=10,y=6=−3×6+2×10=−18+20=2

Since are is m cm2dA/dt=2 cm2/min

Hence are is increasing at the rate of 2 cm2/min
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Answered on 27/01/2022 Learn Applications of Derivatives

Pugazhendhi V

IT Technical Executive with 4 years of experience

Solution: Given, f(x)=4x3−6x2−72x+30 implies f/(x)=12x2−12x−72 a) for strictly increasing f/(x)>0 12x2−12x−72>0 12(x2−x−6)>0 x2−x−6>0 x2−3x+2x−6>0 x(x−3)+2(x−3)>0 (x−3)(x+2)>0 xε(−∞,−2)U(3,∞) b)... read more

Solution:

## Given,

f(x)=4x3−6x2−72x+30
implies f/(x)=12x2−12x−72
a) for strictly increasing
f/(x)>0
12x2−12x−72>0
12(x2−x−6)>0
x2−x−6>0
x2−3x+2x−6>0
x(x−3)+2(x−3)>0
(x−3)(x+2)>0
xε(−∞,−2)U(3,∞)
b) for strictly decreasing
f/(x)<0
12x2−12x−72<0
12(x2−x−6)<0
x2−x−6<0
x2−3x+2x−6<0
x(x−3)+2(x−3)<0
(x−3)(x+2)<0
xε(−2,3)
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Answered on 27/01/2022 Learn Applications of Derivatives

Pugazhendhi V

IT Technical Executive with 4 years of experience

Solution: Let V and S be the volume and surface area of a cube of side x cm respectively. Given dtdV=9 cm3/sec We require x=10cm Now V=x3 ⟹dtdv=3x2.dtdx ⟹9=3x2.dtdx ⟹dtdV=3x29=x23 Again, S=6x2 ⟹dtds=12.x.dtdx =12x.x23=x36 ⟹x=10cm=1036=3.6cm2/sec read more

Solution:

Let V and S be the volume and surface area of a cube of side x cm respectively.
Given dtdV=9 cm3/sec
We require [dtds]x=10cm

Now V=x3
⟹dtdv=3x2.dtdx
⟹9=3x2.dtdx
⟹dtdV=3x29=x23

Again, S=6x2            [By formula for surface area of a cube]
⟹dtds=12.x.dtdx
=12x.x23=x36
⟹[dtdS]x=10cm=1036=3.6cm2/sec
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Answered on 27/01/2022 Learn Applications of Derivatives

Pugazhendhi V

IT Technical Executive with 4 years of experience

Solution: Given; f(x) = (x + 1)3 (x – 3)3 ⇒ f'(x) = (x + 1)3 3(x – 3)2 + (x – 3)33(x + 1)2 = 3(x + 1)2(x – 3)2(x + 1 + x – 3) = 3(x + 1)2(x – 3)2(2x – 2) = 6(x +1)2 (x – 3)2 (x -1) ⇒ 6(x +1)2 (x – 3)2 (x – 1) = 0 ⇒ x = -1, 1,... read more

Solution:

Given;  f(x) = (x + 1)3 (x – 3)3 ⇒ f'(x) = (x + 1)3 3(x – 3)2 + (x – 3)33(x + 1)2 = 3(x + 1)2(x – 3)2(x + 1 + x – 3) = 3(x + 1)2(x – 3)2(2x – 2) = 6(x +1)2 (x – 3)2 (x -1) ⇒ 6(x +1)2 (x – 3)2 (x – 1) = 0 ⇒ x = -1, 1, 3 The intervals are (-∞, -1), (-1, 1), (1, 3), (3, ∞) f'(-2) = (-2 – 1) < 0 ∴ Strictly decreasing in (-∞, -1) f'(0) = (0 – 1) < 0 ∴ Strictly decreasing in (-1, 1) f'(2) = (2 – 1) > 0 ∴ Strictly increasing in (1, 3) f'(4) = (4 – 1) > 0 ∴ Strictly increasing in (3, ∞)

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Answered on 27/01/2022 Learn Applications of Derivatives

Pugazhendhi V

IT Technical Executive with 4 years of experience

Solution: r=9cm Δ r=0.03cm V sphere =34πr3 ΔV=drdvΔr ΔV=34π.3r2.Δr ΔV=34×722×3×9×9×1000.03=30.548 cm read more

Solution:

r=9cm Δ r=0.03cm

V sphere =34πr3  ΔV=drdvΔr

ΔV=34π.3r2.Δr

ΔV=34×722×3×9×9×1000.03=30.548 cm
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Answered on 06 Apr Learn Applications of Derivatives

Let's denote: xx as the distance from the base of the ladder to the wall. yy as the height of the ladder on the wall. According to the Pythagorean theorem, for any position of the ladder, we have x2+y2=52x2+y2=52, since the ladder is always 5 meters long. Now, we differentiate both sides of this... read more

Let's denote:

• xx as the distance from the base of the ladder to the wall.
• yy as the height of the ladder on the wall.

According to the Pythagorean theorem, for any position of the ladder, we have x2+y2=52x2+y2=52, since the ladder is always 5 meters long.

Now, we differentiate both sides of this equation with respect to time tt, considering that xx is changing with time: ddt(x2+y2)=ddt(52)dtd(x2+y2)=dtd(52) 2xdxdt+2ydydt=02xdtdx+2ydtdy=0

Given that the ladder is being pulled away from the wall at a rate of dxdt=2dtdx=2 cm/s, and we want to find dydtdtdy when x=4x=4 m, we can plug in these values into the equation:

2(4)(2)+2ydydt=02(4)(2)+2ydtdy=0 16+4ydydt=016+4ydtdy=0 4ydydt=−164ydtdy=−16 dydt=−164ydtdy=−4y16

Now, we need to find yy when x=4x=4. Using the Pythagorean theorem: 42+y2=5242+y2=52 16+y2=2516+y2=25 y2=9y2=9 y=3y=3

Now, we can find dydtdtdy when y=3y=3: dydt=−164⋅3dtdy=−4⋅316 dydt=−43dtdy=−34

So, the height of the ladder on the wall is decreasing at a rate of 4334 meters per second when the foot of the ladder is 4 meters away from the wall.

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Answered on 27/01/2022 Learn Applications of Derivatives

Pugazhendhi V

IT Technical Executive with 4 years of experience

Solution: Step 1: Differentiating given curve to find rate of change of y w.r.t t. The equation of the curve is given as, 6y=x3+2...........(1) Differentiating (1) both sides w.r.t.t ⇒6dtdy=3x2dtdx+0 ⇒2dtdy=x2dtdx.....(2) Step 2: Solve to find the required point on the given curve. ... read more

Solution:

Step 1: Differentiating given curve to find rate of change of y w.r.t t.

## The equation of the curve is given as,   6y=x3+2...........(1)

Differentiating (1) both sides w.r.t.t

⇒6dtdy=3x2dtdx+0

⇒2dtdy=x2dtdx.....(2)

Step 2: Solve to find the required point on the given curve.

∵It is given that, y coordinate is changing 8 times as fast as the x-coordiante

⇒dtdy=8dtdx..............(3),

Thus we have from (2) and (3),

2(8dtdx)=x2dtdx

⇒16dtdx=x2dtdx

⇒(x2−16)dtdx=0

⇒x2−16=0⇒x=±4                  [∵dtdx=0]

When ,x=4⇒y=643+2=666=11                 [from(1)]

when ,x=(−4)⇒y=6(−4)3+2=−662=−331                     [from(1)]

Hence, the points required on the curve are (4, 11)and (−4,3−31).
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Answered on 27/01/2022 Learn Applications of Derivatives

Pugazhendhi V

IT Technical Executive with 4 years of experience

Solution: The equation of the line is 5y−15x=13. Slope of the line =3 If a tangent is perpendicular to the line 5y−15x=13, then the slope of the tangent is Slope of the line−1=3−1. ⇒dxdy=2x−2=3−1 ⇒2x=3−1+2 ⇒2x=35 ⇒x=65 Now, at x=65 ⇒y=3625−610+7=3625−60+252=36217 Thus,... read more

Solution:

## The equation of the line is 5y−15x=13.

Slope of the line =3

If a tangent is perpendicular to the line 5y−15x=13,

then the slope of the tangent is Slope of the line−1=3−1.

⇒dxdy=2x−2=3−1

⇒2x=3−1+2

⇒2x=35

⇒x=65

Now, at  x=65

⇒y=3625−610+7=3625−60+252=36217

Thus, the equation of the tangent passing through (65,36217) is given by,

(y−36217)=−31(x−65)

⇒=3636y−217−181(6x−5)

⇒36y−217=−2(6x−5)

⇒36y−217=−12x+10

⇒36y+12x−227=0

Hence, the equation of the tangent line to the given curve
(which is perpendicular to line 5y−15x=13) is 36y+12x−227=0
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Answered on 06 Apr Learn Applications of Derivatives

To find the marginal revenue (MR) when 17 units are produced, we first need to find the derivative of the revenue function R(x)R(x) with respect to xx. The marginal revenue is the rate of change of total revenue with respect to the number of units produced. Given that R(x)=13x2+26x+15R(x)=13x2+26x+15,... read more

To find the marginal revenue (MR) when 17 units are produced, we first need to find the derivative of the revenue function R(x)R(x) with respect to xx. The marginal revenue is the rate of change of total revenue with respect to the number of units produced.

Given that R(x)=13x2+26x+15R(x)=13x2+26x+15, we find the derivative R′(x)R(x) and evaluate it at x=17x=17.

First, let's find R′(x)R(x): R′(x)=dRdx=ddx(13x2+26x+15)R(x)=dxdR=dxd(13x2+26x+15)

Using the power rule of differentiation: R′(x)=26x+26R(x)=26x+26

Now, we evaluate R′(x)R(x) at x=17x=17: R′(17)=26(17)+26R(17)=26(17)+26 R′(17)=442+26R(17)=442+26 R′(17)=468R(17)=468

So, the marginal revenue when 17 units are produced is 468468 Rs/unit.

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Answered on 27/01/2022 Learn Applications of Derivatives

Pugazhendhi V

IT Technical Executive with 4 years of experience

Solution: y=(x−2)2 dxdy=dxd((x−2)2)=2(x−2) ∴ slope of tangent =2x−4 Slope of line joining (2,0) and (4,4) =4−24−0=2 The tangent is parallel to this line ∴ their slopes are equal 2x−4=2 ⇒2x=6 ∴x=3 and y=(3−2)2=1 Thus the... read more

Solution:

y=(x−2)2
dxdy=dxd((x−2)2)=2(x−2)
∴ slope of tangent =2x−4
Slope of line joining (2,0) and (4,4) =4−24−0=2
The tangent is parallel to this line
∴ their slopes are equal
2x−4=2 ⇒2x=6
∴x=3
and y=(3−2)2=1
Thus the point is (3,1)
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