application of differentiation question (1 Viewer)

[AGB]

can someone please help me with this?

an isosceles triangle has equal sides of length 10cm. the angle @ between these equal sides is increasing at the rate of 3 degress per minute. show that the area of the triangle is increasing at (5sqrt3.pi)/12 cm^2 per minute at the instant when @ = 30 degrees

thanks in advance

 

[wogboy]

The area of the triangle is:

A = (1/2)*a*b*sin@
= 50*sin@ cm^2

so dA/d@ = 50*cos@
but you know:
d@/dt = 3 deg/min
= pi/60 rad/min
(remember to always use radians in questions like these)

so dA/dt = dA/d@ * d@/dt
= (5/6)*pi*cos@ cm^2 / min

sub in @ = 30 deg = pi/6 rad
dA/dt = (5/6)*pi*cos(pi/6) cm^2 /min
= (5/12)*sqrt(3)*pi cm^2 / min

 

[AGB]

thank you