Related Rates Math?

Question

In a right triangle, an angle is increasing at a constant rate of 15/26 radians per minute. At what rate is the area of the triangle increasing, in square units per minute, when the opposite side is 24 units and the adjacent side is 10? Please explain how this is done. Thanks.

Answer 1

Let the angle increasing at a rate of 15/26 rad per minute be θ
Then tanθ = (length of opposite side) / (length of adjacent side)
= y/x
So y = xtanθ

Now area of triangle (A) = ½xy
= ½x²tanθ

So dA/dt = dA/dθ . dθ/dt
= ½x²sec²θ . dθ/dt

When x = 24 and y = 10, hypotenuse = 26 (26² = 10² + 24²)
So cosθ = 10/26 = 5/13 and thus secθ = 13/5)

So dA/dt = ½(10)²*(13/5)* 15/26 (as dθ/dt = 15/26 ... given)
= 195 units²/min

Answer 2

yay, finally someone who has something more than algebra to ask.
so you write the equation for what you want to find the rate of (this process applies to every process of related rates).
(well, first it depends on which angle is getting bigger. so i'ma gonna assume that the 24 side is not the largest side of the triangle)
so you know from pythagreon pythagrion, whatever... that the third side of the triangle must be 26.

now,
Area= 1/2 * a*b* Sin C.
what this means is that Area = 1/2 times lengh of two adjacent sides (adjacent to the vertex or angle you're looking at) times the angle that you're looking at.
so we know that

Area= 1/2 *10*26 * sin C.
Area= 130 Sin C
now we take the derivative.
d (Area) / dt = 130 * (cos C) * d(C)/dt

that translates into "the rate the area is changing is equal to 130 times the cosine of the angle (at that instant) times how fast the angle is changing."

so
remember that we know C.= 67.38 degrees ( becausesine C = 24/26)

so now all you have to do is multiply-
d(area)/dt = 130* cos( 67.38) * (15/26 radians per sec).

yeah?
just remember.
find the equation for what you wanna fine the rate of, and then take the derivative, then plug in stuff you know.


(hmm.. hope i didn't make any mistakes >.<)

Answer 3

x=opp side y=adj side z=angle A=area
x/y=tanz x=ytanz

A=xy/2= y^2tan(z)/2

dA/dt= (1/2)[2ytan(z)dy/dt+
y^2sec^2(z)dz/dt]

we know 'y' and dz/dt since they were given. z=tan^-1(x/y)= tan^-1(24/10) so z is known

dy/dt=0 I believe since opening the angle does not change the lenght of 'y'.
Now that all of the variable are known plug them in to get dA/dt