Thread: Max/Min and Related Rates Question...

1. Max/Min and Related Rates Question...

#1 If a and b are lengths of two sides of a triangle, and theta the measure of hte included angle, the area A of hte triangle is A = (1/2)absin(theta). How is dA/dt related to da/dt, db/dt, and dtheta/dt?

My work:
A= (1/2)absin(theta)
dA/dt = ((1/2)(ab))'sin(theta) + (1/2)ab(sin(theta))'
dA/dt = ((1/2)(dA/dt)b + (1/2)a(db/dt))sin(theta) + (1/2)abcos(theta)(theta)'

I think there's something wrong either in the power rule or the (theta)'.

#2 Why do we use the formula
D^2 = x^2 + y^2 in the following question?...
A point moves smoothly along the curve y = x^(3/2) in the first quadrant in such a way that its distance from the origin increases at the constant rate of 11 units per second. Find dx/dt when x = 3.
I know the answer is dx/dt = 4 units per second. I just want to know why we use that formula, and other potential scenerios of when we might use it.

2. For question 1 , it is not mentioned that if a, and $\theta$.
So letrs assume that all three are dependent upon time.
$
dA/dt = d(1/2ab\sin\theta)$
/dt

Now apply multiplication rule of differentiation.

$dA/dt = 1/2ab * d\sin\theta/dt + 1/2\sin\theta * d(ab)/dt$
= $1/2ab *\cos\theta d\theta/dt + 1/2 \sin\theta * (a*db/dt + b*da/dt)$

Simplify and answer will be in terms of da/dt, db/dt and $d\theta/dt$

according to question rate change of distance from origin is 11 unit per second

$=>d( \sqrt(x^2 - y^2))/dt = 11$
$=> d\sqrt(x^2 - (x^(3/2))^2)/dt = 11^2$ Since point lies on given curve $y = x^(3/2)$

solve for dx/dt