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

10. A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall ?

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Vishal Baghel | Contributor-Level 10

4 months ago

Since, the bottom of ground is increasing with time t,

$\frac{d x}{d t}$ = 2cm/s

From fig, Δ ABC, by Pythagorastheorem

AB² + BC² = AC²

$\Rightarrow$ x² + y² = 5²

x² + y² = 25 ____ (1)

Differentiating eqⁿ (1) w. r. t. time t we get,

$\frac{d}{d t} (x^{2} + y^{2}) = \frac{d}{d t} (25)$

$\Rightarrow 2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0 .$

$\Rightarrow 2 x \times 2 + 2 y \frac{2 y}{d y} = 0$

$\Rightarrow \frac{d y}{d t} = - \frac{2 x}{y}$ m/s

When x = 4m, the rate at which its height on the wall decreases is

$\frac{d y}{d t} = - \frac{2 \times 4}{3}$ ${\begin{cases} ∴ 4^{2} + y^{2} = 5^{2} \\ \Rightarrow y^{2} = 25 - 16 \\ \Rightarrow y √9 (L engthcan'tbenegetive) \\ \Rightarrow y = 3 \end{cases}$

$\Rightarrow \frac{d y}{d t} = - \frac{8}{3}$ room

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If the normal to the curve y(x) = ∫(from 2 to x) (2t² - 15t + 10)dt at a point (a, b) is parallel to the line x + 3y = -5, a > 1, then the value of |a + 6b| is equal to..........

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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).
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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
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a = 1/2 Rejected as a > 1. So a = 7.
b = ∫? (2t² - 15t + 10)dt = [2t³/3 - 15t²/2 + 10t] from 0 to 7.
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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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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

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

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

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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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10. A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall ?

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