this is a relate rates problem in calculus
in a right triangle, the hypotenuse is 5 and the vertical side is x and the angle is theta.
so if theta increases at a constant rate of 3 radians per minute, at what rate is x increasing in units per minute when x equals 3 units?
d(theta)/dt=3 rad/min
so if you draw the triangle, you get:
sin(theta)=x/5::::theta=u
take derivative:
cos(u(du/dt)=(1/5)dx/dt
i solve for dx/dt and i'm assuming the theta is 45 degrees because it's right triange. but the problem asks for when x=3 but the x is gone so what do i do?
no!!!
you had x = 5sin(theta), right?
now you have to differentiate that with respect to t, because it is a rate
dx/dt = 5cos(theta) d(theta)/dt
only now do you consider the case when x=3
when x=3, r=5, then the third side =4
and cos(theta)= 4/5
also you were given that d(theta)/dt = 3 rad/min
so.....
dx/dt = 5(4/5)(3) units/min
=15 units/min
5 x (4/5) x 3 = 12
Thank you!!
To start, we have x = 5sin(theta). Now, we need to differentiate this equation with respect to time, t, since we are looking for the rate at which x is increasing.
Differentiating x = 5sin(theta) with respect to t, we get
dx/dt = 5cos(theta) * d(theta)/dt
Now, we can consider the case when x = 3. Given that the hypotenuse is 5 and the vertical side is x, we can use the Pythagorean theorem to find the length of the third side.
When x = 3, the third side (let's call it r) can be found using the Pythagorean theorem:
r^2 = 5^2 - 3^2
r^2 = 25 - 9
r^2 = 16
r = 4
Now, we need to find the value of cos(theta) when x = 3.
cos(theta) = r/5 = 4/5
Finally, we were given that d(theta)/dt = 3 rad/min.
Putting it all together, we can calculate dx/dt:
dx/dt = 5cos(theta) * d(theta)/dt
dx/dt = 5(4/5)(3)
dx/dt = 15 units/min
To solve this related rates problem, the first step is to express the relationship between the variables. In this case, the right triangle gives the relationship between the hypotenuse and the vertical side: sin(theta) = x/5.
Taking the derivative with respect to time (t), you get: cos(theta) * d(theta)/dt = (1/5) * dx/dt.
Now, you need to substitute the given values into the equation. When x = 3 units, the hypotenuse (5) and the third side of the triangle are fixed. From the triangle, you can determine that the third side is 4 units and cos(theta) = 4/5.
You were given that d(theta)/dt = 3 rad/min, so now you can substitute all the values into the equation:
dx/dt = 5 * (4/5) * 3 units/min
= 15 units/min
Therefore, when x = 3 units, the vertical side is increasing at a rate of 15 units per minute.