The two sides of a isosceles triangle have a fixed lenght of 14 cm. The opposite angle at the base of the triangle increases by 0.3 rad/min.
a) what is the growth rate dx/dt of the base of the triangle when the opposite angle mesures 1.6 rad? (Use the law of sin)
b) What is the growth rate dA/dt of the the triangles area when the opposite angle at the base mesures 1.6 rad?
again I will show you how to do one of them. T is the angle between equal sides. x is opposite side.
sin T/x = sin [(pi -T)/2]/14
x sin [(pi-T)/2] = 14 sin T
x cos[(pi-T)/2] (-dT/dt)+sin[(pi-T)/2](dx/dt) =14 cos T dT/dt
here dT/dt = .3 rad/min
starting:
T = 1.6 (about a right angle)
sin T = 1.00
cos T = 0
T is close enough to 90 degrees
(pi-T)/2 = (3.14-1.6)/2 = .77
cos .77 = .717 call it .707 (45 deg)
sin .77 = .697 call it .707 (45 deg)
so
x sin [(pi-T)/2] = 14 sin T
x = 14 (1)/.707 = 19.8
x cos[(pi-T)/2] (-dT/dt)+sin[(pi-T)/2](dx/dt) =14 cos T dT/dt
19.8(.707)(-.3)+.707dx/dt = 0
dx/dt = 5.94 cm/min
tnx, I udnerstand how to do them now, cauz I have an exam tomorrow, and I just didn';t understand the way my teacher showed me, tnx again!
You are welcome. Always check my arithmetic !
To solve these problems, we can use the law of sines, which states that the ratio of the length of a side of a triangle to the sine of its opposite angle is constant.
a) We need to find the growth rate of the base (dx/dt) when the opposite angle measures 1.6 rad. Let's assume the base of the triangle is denoted by x.
Using the law of sines, we have:
sin(1.6) / 14 = sin(angle opposite to x) / x
Now, we need to differentiate both sides with respect to time t using implicit differentiation. Remember that the derivative of sin(x) with respect to x is cos(x), and the derivative of x with respect to t is dx/dt.
cos(1.6) (1/14) = cos(angle opposite to x) * (dx/dt) / x
Solving for dx/dt, we have:
dx/dt = (cos(1.6) / 14) * x / cos(angle opposite to x)
Substituting cos(angle opposite to x) = cos(1.6) and substituting x = 14 cm (as given in the problem), we get:
dx/dt = (cos(1.6) / 14) * 14 / cos(1.6)
Simplifying further, we find:
dx/dt = cos(1.6)
Therefore, the growth rate of the base of the triangle is dx/dt = cos(1.6) cm/min.
b) To find the growth rate of the triangle's area (dA/dt) when the opposite angle at the base measures 1.6 rad, we can use the formula for the area of a triangle:
A = (1/2) * (side length 1) * (side length 2) * sin(opposite angle)
Let's assume side length 1 and side length 2 are both 14 cm (as given in the problem). So, the area is:
A = (1/2) * 14 * 14 * sin(1.6)
Now we need to differentiate both sides with respect to time t using implicit differentiation. The derivative of sin(x) with respect to x is cos(x), and the derivative of x with respect to t is dx/dt.
dA/dt = (1/2) * 14 * 14 * cos(1.6) * (d(1.6)/dt)
Substituting d(1.6)/dt = 0.3 (as given in the problem), we have:
dA/dt = (1/2) * 14 * 14 * cos(1.6) * 0.3
Simplifying further, we find:
dA/dt = 29.2743 * cos(1.6)
Therefore, the growth rate of the triangle's area is dA/dt = 29.2743 * cos(1.6) square cm/min.