Question

Two sides of a triangle have lengths 13 m and 17 m. The angle between them is increasing at a rate of 2°/min. How fast (in m/min) is the length of the third side increasing when the angle between the sides of fixed length is 60°? [Hint: Use the Law of Cosines.] (Round your answer to three decimal places.)

Answer 1

remember, we see the limits of changes (the change rate) focused on a single point via the first derivative of a function.

the 3 sides of a triangle are related to each other via the law of cosine (which I always call the generalized principle of Pythagoras) :

c² = a² + b² - 2ab×cos(C)

with C being the angle opposite of the side c. this works now for every side of the triangle. not just for a Hypotenuse in a right-angled triangle.

c = sqrt(a² + b² - 2ab×cos(C))

C is the changing angle (2°/ minute).

c is the third growing side of the triangle.

we want to check about the change of C and its effects on c.

dC/dt = 2°/minute, and we need this as length. so, we need to convert this to radians.

1° = pi/180

2° = 2pi/180 = pi/90

dc/dt = d(sqrt(a² + b² - 2ab×cos(C)))/dC × dC/dt=

= d(sqrt(13² + 17² - 2×13×17×cos(C)))/dC × dC/dt =

= d(sqrt(169 + 289 - 442×cos(C)))/dC × dC/dt =

= d(sqrt(458 - 442×cos(C)))/dC × dC/dt

remember,

(f(x)^n)' = n × f(x)^(n-1) × f'(x)

as a square root is nothing else than an 1/2 exponent.

d(sqrt(458 - 442×cos(C)))/dC =

= d((458 - 442×cos(C))^½)/dC =

= ½ × (458 - 442×cos(C))^-½ × d(458 - 442×cos(C))/dC

we also remember that a negative exponent means 1/...

½ × (458 - 442×cos(C))^-½ =

= ½ × 1/sqrt(458 - 442×cos(C)) =

= 1/(2×sqrt(458 - 442×cos(C)))

cos(x)' = -sin(x)

d(458 - 442×cos(C))/dC = 0 - 442×-sin(C)=

= 442×sin(C)

so, when combining both factors again, we get

dc/dt =

= 1/(2×sqrt(458 - 442×cos(C))) × 442×sin(C) × dC/dt =

= 1/sqrt(458 - 442×cos(C)) × 221×sin(C) × dC/dt =

= 1/sqrt(458 - 442×cos(C)) × 221×sin(C) × pi/90 / min

at C = 60° we get

dc/dt = 1/sqrt(458 - 442×cos(60)) × 221×sin(60) × pi/90 =

= 1/sqrt(458 - 442×½) × 221×sqrt(3)/2 × pi/90 =

= 1/sqrt(458 - 221) × 221×sqrt(3)/2 × pi/90 =

= 1/sqrt(237) × 221×sqrt(3)/2 × pi/90 =

= 1/sqrt(3×79) × 221×sqrt(3)/2 × pi/90 =

= 1/(sqrt(3)×sqrt(79)) × 221×sqrt(3)/2 × pi/90 =

= 221pi / (sqrt(79)×2×90) = 0.43396639... m/min ≈

≈ 0.434 m/min

the third side increases at that moment (angle = 60°) by 0.434 meters/minute.

Answer 2

I don’t know whether it’s correct but, this is what I got

2 degrees - 1 min
60 degrees - x mins

Find x and you get 30 mins.

Then you find the third side:

Let the third side be y

y2 = (13 squared) + 17 squared - 2(13 * 17) cos 60
y2 = 169 + 289 - 442cos60
y2 = sqr root (237)
y = 15.395cm to 3 d.p

So if

60 degrees - 15.395
2 degrees - x

Find x
You get 0.513 cm


(15.395 - 0.513) / (30 - 2)
14.882/28
0.532 m per min (to 3d.p)