Let x be the length of each of the equal side | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The area $A$ of an isosceles triangle with equal sides $x$ and included angle $	heta$ is given by $A = \frac{1}{2} x^2 \sin 	heta$.To find the r

Vedclass · Accepted Answer

$\sqrt{2}\left(\frac{\pi}{5}+\frac{1}{2}ight) 	ext{ m}^2/	ext{hour}$

Vedclass · Answer

(B) The area $A$ of an isosceles triangle with equal sides $x$ and included angle $	heta$ is given by $A = \frac{1}{2} x^2 \sin 	heta$.To find the rate of change of the area,we differentiate $A$ with respect to time $t$ using the product rule:$\frac{dA}{dt} = \frac{1}{2} \left( 2x \frac{dx}{dt} \sin 	heta + x^2 \cos 	heta \frac{d	heta}{dt} ight) = x \frac{dx}{dt} \sin 	heta + \frac{1}{2} x^2 \cos 	heta \frac{d	heta}{dt}$.Given values: $x = 12 	ext{ m}$,$	heta = \frac{\pi}{4}$,$\frac{dx}{dt} = \frac{1}{12} 	ext{ m/h}$,and $\frac{d	heta}{dt} = \frac{\pi}{180} 	ext{ rad/h}$.Substituting these values into the derivative:$\frac{dA}{dt} = (12) \left( \frac{1}{12} ight) \sin \left( \frac{\pi}{4} ight) + \frac{1}{2} (12)^2 \cos \left( \frac{\pi}{4} ight) \left( \frac{\pi}{180} ight)$.$\frac{dA}{dt} = (1) \left( \frac{1}{\sqrt{2}} ight) + \frac{1}{2} (144) \left( \frac{1}{\sqrt{2}} ight) \left( \frac{\pi}{180} ight)$.$\frac{dA}{dt} = \frac{1}{\sqrt{2}} + \frac{72}{\sqrt{2}} \left( \frac{\pi}{180} ight) = \frac{1}{\sqrt{2}} + \frac{2\pi}{5\sqrt{2}} = \frac{1}{\sqrt{2}} \left( 1 + \frac{2\pi}{5} ight)$.Note: The provided options seem to contain a typo in the coefficient. Re-evaluating the expression: $\frac{1}{\sqrt{2}} (1 + \frac{2\pi}{5}) = \frac{\sqrt{2}}{2} (1 + \frac{2\pi}{5}) = \sqrt{2} (\frac{1}{2} + \frac{\pi}{5})$.Thus,the correct option is $B$.

Similar Questions

Two particles $A$ and $B$ move from rest along a straight line with constant accelerations $f$ and $f'$ respectively. If $A$ takes $m$ seconds more than that of $B$ and describes $n$ units more than that of $B$ in acquiring the same velocity, then:

The equation of motion of a particle moving along a straight line is $s = 2t^{3} - 9t^{2} + 12t$,where the units of $s$ and $t$ are centimetre and second. The acceleration of the particle will be zero after: (in $s$)

The rate of change of the area of a circle with respect to its radius $r$ at $r = 2 \text{ cm}$ is:

$A$ kite is $120 \ m$ high and $130 \ m$ of string is out. If the kite is moving away horizontally at the rate of $39 \ m/sec$,then the rate at which the string is being paid out is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module