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Limits and Derivatives > Unit III: Calculus > Learn Applications of Derivatives

Applications of Derivatives

Applications of Derivatives relates to CBSE - Class 11/Mathematics/Limits and Derivatives/Unit III: Calculus

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Answered on 27/01/2022 Learn CBSE - Class 11/Mathematics/Limits and Derivatives/Unit III: Calculus/Applications of Derivatives

The length x of a rectangle is decreasing at the rate of 3 cm/ mint and the width y is increasing at...
The length x of a rectangle is decreasing at the rate of 3 cm/ mint and the width y is increasing at the rate of 2cm/min. when x = 10cm and y = 6cm, find the ratio of change of (a) the perimeter (b) the area of the rectangle

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 CBSE - Class 11/Mathematics/Limits and Derivatives/Unit III: Calculus/Applications of Derivatives

Find the interval in which the function given by f(x) = 4x3 -- 6x2 -- 72x + 30 is (a) strictly increasing...
Find the interval in which the function given by f(x) = 4x3 – 6x2 – 72x + 30 is (a) strictly increasing (b) strictly decreasing.

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 CBSE - Class 11/Mathematics/Limits and Derivatives/Unit III: Calculus/Applications of Derivatives

The volume of a cube is increasing at a rate of 9cm3/s. How fast is the surface area increasing when...
The volume of a cube is increasing at a rate of 9cm3/s. How fast is the surface area increasing when the length of on edge is 10cm?

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 CBSE - Class 11/Mathematics/Limits and Derivatives/Unit III: Calculus/Applications of Derivatives

Find the interval in which the function is strictly increasing and decreasing. (x+1)3 (x-3)3

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 CBSE - Class 11/Mathematics/Limits and Derivatives/Unit III: Calculus/Applications of Derivatives

IF the radius of a sphere is measured as 9cm with an error of 0.03cm, then find the approximate error...
IF the radius of a sphere is measured as 9cm with an error of 0.03cm, then find the approximate error in calculating its volume.

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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