ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.9

Students can cross-reference their work with ML Aggarwal Class 12 ISC Solutions Chapter 7 Applications of Derivatives Ex 7.9 to ensure accuracy.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.9

Typical Problems:

Question 1.
The radius of the base of right circular cone is increasing at the rate of 3 cm per minute and the altitude is decreasing at the rate of 4 cm per minute. Find the rate of change of total surface area of the cone when its radius is 7cm and altitude is 24 cm.
Solution:

Let r be the radius and h be the height of right circular cone.
Let S = Total surface area of right circular cone = πr² + πrl
= πr² + πr \(\sqrt{r^2+h^2}\) …..(1)
[where l = slant height of cone = \(\sqrt{r^2+h^2}\)]
Diff. (1) w.r.t. t ; we get
\(\frac{d \mathrm{~S}}{d t}=2 \pi r \frac{d r}{d t}+\pi\left[\sqrt{r^2+h^2} \frac{d r}{d t}+\frac{r}{2 \sqrt{r^2+h^2}}\left[2 r \frac{d r}{d t}+2 h \frac{d h}{d t}\right]\right]\)
= \(2 \pi r \frac{d r}{d t}+\pi\left[\sqrt{r^2+h^2} \frac{d r}{d t}+\frac{r}{\sqrt{r^2+h^2}}\left\{r \frac{d r}{d t}+h \frac{d h}{d t}\right\}\right]\)
Given \(\frac{d r}{d t}\) = 3 cm / min.
\(\frac{d h}{d t}\) = – 4 cm/min ;
r = 7 cm and h = 24 cm
∴ from (1) ; we have
\(\frac{d S}{d t}\) = 2π × 7 × 3 + π \(\left[\sqrt{7^2+24^2} \times 3+\frac{7}{\sqrt{7^2+24^2}}\{7 \times 3+24(-4)\}\right]\)
= 42π + π [25 × 3 + \(\frac{7}{25}\) (21 – 96)]
= 42π + π [75 – 21]
= 96π cm² /min.

Question 2.
Find the equation of normal to the curve y = x + \(\frac{2}{x}\) at the point where abscissa is 2. If this normal meets the axes in points A and B find the length of AB.
Solution:
Given eqn. of curve y = x + \(\frac{2}{x}\) …………..(1)
given point on the curve (1) whose abscissa is 2.
Thus, x = 2
∴ from (1) ;
y = 2 + \(\frac{2}{2}\) = 3
∴ point on given curve becomes (2, 3).
Differentiating eqn. (1) w.r.t. x; we have
\(\frac{d y}{d x}\) = 1 – \(\frac{2}{x^2}\)
∴ Slope of normal to given curve at point (2, 3) = \(-\frac{1}{\left(\frac{d y}{d x}\right)_{(2,3)}}\)
= \( \frac{-1}{\left(1-\frac{2}{4}\right)}=\frac{-1}{\frac{1}{2}}\) = – 2
;. required eqn. of normal to given curve at point (2, 3) is given by
y – 3 = – 2(x – 2)
⇒ 2x + y = 7 ………..(2)
Now let eqn. (2) meets x-axis at A ¡e. y = 0
∴ from (2);
x = \(\frac{7}{2}\)
∴ coordinates of A are (\(\frac{7}{2}\), 0)
Let eqn. (2) meets y-axis at B i.e. x = O
∴ from (2) ;
y = 7
Thus coordinates of point B are (0, 7)
∴ |AB| = length of AB
= \(\sqrt{\left(0-\frac{7}{2}\right)^2+(7-0)^2}\)
= \(\sqrt{\frac{49}{4}+49}\)
= \(\frac{7}{2}\) √5 units.

Question 3.
Show that for all a ≥ 1, the function f defined by f(x) = √3 sin x – cos x – 2ax + b is decreasing on R.
Solution:
Given f(x) = √3 sin x – cos x – 2ax + b
putting √3 = r cos α ; – 1 = r sin α
On squaring and adding; we have
r = 2 and on dividing; we have
tan α = – \(\frac{1}{\sqrt{3}}\)
= tan (- \(\frac{\pi}{6}\))
⇒ α = – \(\frac{\pi}{6}\)
∴f(x) = 2 sin (x – \(\frac{\pi}{6}\)) – 2ax + b
Diff. both sides w.r.t. x, we have
f’ (x) = 2 cos (x – \(\frac{\pi}{6}\)) – 2ax + b
Diff. both sides w.r.t. x, we have
f'(x) = 2 cos (x – \(\frac{\pi}{6}\)) – 2a
Now – 1 ≤ cos (x – \(\frac{\pi}{6}\)) ≤ 1
⇒ – 2 ≤ 2 cos (x – \(\frac{\pi}{6}\)) ≤ 2
⇒ – 2 – 2a ≤ 2 cos (x – \(\frac{\pi}{6}\)) – 2a ≤ 2 – 2a ≤ 0
[if a ≥ 1
⇒ 2a ≥ 2
⇒ 2 – 2a ≤ 0]
⇒ f'(x) ≤ 0
Thus f(x) be a decreasing function if a ≥ 1.

Question 4.
Find the greatest and the least value of f(x) = x² log x in [1, e].
Solution:
Given f(x) = x² log x in [ 1, e]
On differentiating both sides, w.r.t. x, we have
f’ (x) = x² × \(\frac{1}{x}\) + (log x) 2x
= x + 2x log x
= x [ 1 + 2 log x]
For critical points ;
f’ (x) = 0
⇒ x (1 + 2 log x) = 0
⇒ x = 0, e^{– 1/2}
but x ∈ e [1, e]
Since 0, e^{– 1/2} ∉ [1, e]
Thus only critical points are 1 and e.
f(1) = 1 log 1 = 0
f(e) = e² log e = e²
∴ greatest value of f(x) = e²
and it attains at x = e
and least value of f(x) = 0
and it attains at x = 0.

Question 5.
A lighthouse is located at A, 2 km off share from the nearest point O on a straight beach and a shop is located B on the beach at a distance of 4 km from O. If the housekeeper can row at the speed of 4 km/h and can walk at the rate of 6 km/h. Where should be plane to reach the shore so as to cover the distance to the shop in the least possible time?
Solution:
Let C be the point between O and B and the man should reach the shore at the point C.
Let OC = x km.
The distance rowed = AC = \(\sqrt{2^2+x^2}\) km.
Let t₁ = rowing time
= distance/speed
⇒ t₁ = \(\frac{\sqrt{2^2+x^2}}{4}\) hr

The distance walked = CB = (4 – x) km
∴ t₂ = walking time = \(\frac{4-x}{6}\) hr
Thus total time required T = t₁ + t₂
∴ T = \(\frac{\sqrt{4+x^2}}{4}+\frac{4-x}{6}\)
Diff. both sides w.r.t. x, we have

On squaring; we have
36x² = 16 (4 + x)
⇒ 9x² = 4 (4 + x²)
⇒ 5x² = 16
⇒ x = ± \(\frac{4}{\sqrt{5}}\) [but x > 0]
∴ x = \(\frac{4}{\sqrt{5}}\)
at x = \(\frac{4}{\sqrt{5}}\) ;
\(\frac{d^2 \mathrm{~T}}{d x^2}=\frac{1}{\left(4+\frac{16}{5}\right)^{3 / 2}}\)
= \(\frac{5 \sqrt{5}}{216}\) > 0
Thus T is minimise for x = \(\frac{4}{\sqrt{5}}\) km
Hence the man should reach the shore at the point whose distance from the point O = \(\frac{4}{\sqrt{5}}\) km
= \(\sqrt{\frac{32}{10}}\) km
= \(\sqrt{3.2}\) km.

Question 6.
A cone is circumscribed about a sphere of radius R. Show that the volume of the cone is minimum if its height is 4R.
Solution:
Let x cm be the radius of the circumscribed right circular cone
and Let h cm be the height of cone.
Then AM = \(\sqrt{\mathrm{OA}^2-\mathrm{OM}^2}\)
= \(\sqrt{(h-\mathrm{R})^2-\mathrm{R}^2}\)
= \(\sqrt{h^2-2 h \mathrm{R}}\)

Let ∠DAC = α,
In right angled ∆ADC, we have
tan α = \(\frac{\mathrm{DC}}{\mathrm{AD}}=\frac{x}{h}\) ………..(1)
Also in right angled ∆OMA, we have
tan α = \(\frac{\mathrm{OM}}{\mathrm{AM}}=\frac{\mathrm{R}}{\sqrt{h^2-2 h \mathrm{R}}}\)
From eqn. (1) and (2) ; we have
\(\frac{x}{h}=\frac{\mathrm{R}}{\sqrt{h^2-2 h \mathrm{R}}}\)
⇒ x = \(\frac{\mathrm{R} h}{\sqrt{h^2-2 h \mathrm{R}}}\)
Let V be the volume of cone which is to be minimised
Then V = latex]\frac{\pi}{3}[/latex] r²h
= \(\frac{\pi}{3} \frac{\mathrm{R}^2 h^2}{\left(h^2-2 h \mathrm{R}\right)}\) × h
= \(\frac{\pi}{3} \frac{\mathrm{R}^2 h^2}{h-2 \mathrm{R}}\)
Diff. both sides w.r.t. h, we have
\(\frac{d \mathrm{~V}}{d h}=\frac{\pi \mathrm{R}^2}{3}\left[\frac{(h-2 \mathrm{R}) 2 h-h^2}{(h-2 \mathrm{R})^2}\right]\)
= \(\frac{\pi \mathrm{R}^2}{3} \frac{\left(h^2-4 \mathrm{R} h\right)}{(h-2 \mathrm{R})^2}\)
⇒ \(\frac{d \mathrm{~V}}{d h}=\frac{\pi \mathrm{R}^2 h(h-4 \mathrm{R})}{3(h-2 \mathrm{R})^2}\)
For maxima/minima, \(\frac{d V}{d h}\) = 0
⇒ h (4 – 4R) = 0
⇒ h = 4R (∵ h > 0)
When h slightly < 4R
⇒ h – 4R < 0
∴ \(\frac{d V}{dh}\) < 0 When h > 4R slightly
⇒ h – 4R > 0
∴ \(\frac{d V}{dh}\) > 0
Thus \(\frac{d V}{dh}\) changes its sign from -ve to positive (+ve) as we move h <4R slightly to h > 4R slightly.
∴ h = 4R be a point of minima.
Hence volume of cone is minimum when height of cone is 4R.