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Ncert Solutions Maths class 12th Maths Class 12th Application of Derivatives

2. The volume of a cube is increasing at the rate of 8 cm3 /s. How fast is the surface area increasing when the length of an edge is 12 cm?

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Vishal Baghel | Contributor-Level 10

4 months ago

Let x be the length of edge,v be the value and s be the surface area of the cube then,

y = x3.

and S = 6x2, where x is a fxn of time.

Now, $\frac{d Y}{d F} = 8 c m^{\frac{3}{5}}$

$\Rightarrow \frac{d}{d t}$ (x3) = 8

$\frac{d x^{3}}{d x} \cdot \frac{d x}{d t} = 8$ (by chain rule)

$\Rightarrow$ 3x2 $\frac{d x}{d x} = 8 .$

$\frac{d x}{d t} = \frac{8}{3 x^{2} .}$

Now, $\frac{d S}{d t} = \frac{d}{d t}$ (bx2) = $\frac{d (6 x^{2})}{d x} - \frac{d x}{d x}$ = 12x $\frac{8}{3 x^{2}} = \frac{32}{x}$ $c m^{\frac{2}{5}}$ .

When x = 12 cm,

$\frac{d S}{d t} = \frac{32}{12} = \frac{8}{3}$ $c m^{\frac{2}{5}}$ .

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y (x) = ∫? (2t² - 15t + 10)dt
dy/dx = 2x² - 15x + 10.
For tangent at (a, b), slope is m = dx/dy = 1 / (dy/dx) = 1 / (2a² - 15a + 10).
Given slope is -1/3.
2a² - 15a + 10 = -3
2a² - 15a + 13 = 0 (The provided solution has 2a²-15a+7=0, suggesting a different problem or a typo)
Following the image: 2a² - 15a + 7 = 0
(2a - 1) (a - 7) = 0
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b = ∫? (2t² - 15t + 10)dt = [2t³/3 - 15t²/2 + 10t] from 0 to 7.
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If the tangent to the curve, y = f(x) = xlog? x, (x > 0) at a point (c, f(c)) is parallel to the line - segment joining the points (1,0) and (e, e) then c is equal to:

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f' (c) = 1 + lnc = e/ (e-1)
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The lengths of the sides of a triangle are 10 + x2, 10 + x2 and 20 – 2x2. If for x = k, the area of the triangle is maximum, then 3k2 is equal to:

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$C D = \sqrt[]{{(10 + x^{2})}^{2} - {(10 - x^{2})}^{2}} = 2 \sqrt[]{10} | x |$

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$= \frac{1}{2} \times C D \times A B = \frac{1}{2} \times 2 \sqrt[]{10} | x | (20 - 2 x^{2})$

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Let F₁(A,B,C) = $(A \land \sim B) \lor$ $[\sim C \land (A \lor B)] \lor \sim A a n d F_{2} (A, B) = (A \lor B) \lor (B \to \sim A)$ be two logical expressions. Then:

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By truth table

So F₁ (A, B, C) is not a tautology

Now again by truth table

So F₂ (A, B) be a tautology.

The triangle of maximum area that can be inscribed in a given circle of radius ‘r’ is:

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From option let it be isosceles where AB = AC then

$x = \sqrt[]{r^{2} - {(h - r)}^{2}}$

= $\sqrt[]{r^{2} - h^{2} - r^{2} + 2 r h}$

$x = \sqrt[]{2 h r - h^{2}} . . . . . . . . (i)$

Now ar $(Δ A B C) = Δ = \frac{1}{2} B C \times A L$

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then $x = \sqrt[]{2 \times \frac{3 r}{2} \times r - \frac{9 r^{2}}{4}} = \frac{\sqrt[]{3}}{2} r f r o m (i)$

$\Rightarrow B C = \sqrt[]{3} r$

So $A B = \sqrt[]{h^{2} + x^{2}} = \sqrt[]{\frac{9 r^{2}}{4} + \frac{3 r^{2}}{4}} = \sqrt[]{3} r$ .

Hence $Δ$ be equilateral having each side of length $\sqrt[]{3} r .$

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2. The volume of a cube is increasing at the rate of 8 cm3 /s. How fast is the surface area increasing when the length of an edge is 12 cm?

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