A particle on the x-axis is moving to the right at 2 units per second. A second particle is moving down the y-axis at the rate of 3 units per second. At a certain instant the first particle is at the point (5,0) and the second is at the point (0,7). How rapidly is the angle between the x-axis and the line joining the two particles changing at that instant? Are the particles moving towards or away from each other at that instant?
I know the x and y values (x=5, y=7) as well as dx/dt=2 and dy/dt=3. I'm stuck on the formula to use for a changing angle. I know to use implicit differentiation from there and to substitute the values in afterwards. Just not sure of the formula to use. Thanks for any help!
let the position on the x-axis be (x,0)
and the position on the y-axis by (0,y)
the article is moving to the right on the x-axis at 2 units/s ---> dx/dt = 2
the article is moving DOWN the y-axis at 3 units/s
---- dy/dt = -3
according to my diagram, at any moment of t seconds,
tanØ = y/x, where Ø is the angle formed at the x-axis
xtanØ = y
x sec^2 Ø dØ/dt + tanØ dx/dt = dy/dt
when x = 5, y = 7, tanØ = 7/5
sketch a right-angled triangle, the hypotenuse can be found:
r^2 = 7^2 + 5^2
r = √74
secØ = 1/cosØ = √74/5
sec^2 Ø = 74/25
sub in your given:
x sec^2 Ø dØ/dt + tanØ dx/dt = dy/dt
5(74/25) dØ/dt + (7/5)(2) = -3
(74/5) dØ/dt = -3-14/5 = -29/5
dØ/dt = (-29/25)(5/74) = -29/370
the angle is decreasing at 29/370 radians/second
check my arithmetic
Let t=0 at the given instant. The slope of the line is tanθ (with θ measured clockwise from the x-axis)
tanθ = (7-3t)/(5+2t) = 7/5
sec^2θ dθ/dt = -29/(5+2t)^2
74/25 dθ/dt = -29/25
dθ/dt = -29/74
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second method
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The distance z between the points is
x^2+y^2 = z^2
x dx/dt + y dy/dt = z dz/dt
5(2)+7(-3) = √74 dz/dt
dz/dt = -11/√74
so the points are getting closer
tanθ = y/x
sec^2θ θ' = (xy'-yx')/x^2
(74/25)θ' = (5(-3)-7(2))/25
θ' = -29/74
Steve got the same answer in 2 different ways.
I must have made an error in my arithmetic, too lazy to find it
To determine how rapidly the angle between the x-axis and the line joining the two particles is changing, we can use trigonometry.
Let's consider the two particles as A and B, with coordinates (5,0) and (0,7) respectively.
The angle between the x-axis and the line joining A and B can be found using the tangent function. The tangent of an angle is defined as the ratio of the opposite side to the adjacent side. In this case, the opposite side is the y-coordinate difference between A and B (Δy = 7 - 0 = 7), and the adjacent side is the x-coordinate difference between A and B (Δx = 5 - 0 = 5).
So, the tangent of the angle θ (theta) between the x-axis and the line joining A and B is given by: tan(θ) = Δy / Δx = 7 / 5.
To find how rapidly the angle is changing, we need to differentiate this expression with respect to time (t). Since both Δy and Δx are changing with time, we can write:
tan(θ) = Δy / Δx
Differentiating both sides with respect to t using implicit differentiation:
sec²(θ) * dθ/dt = (dΔy/dt * Δx - Δy * dΔx/dt) / Δx²
We already know that dΔy/dt = -3 and dΔx/dt = 2. Substituting these values into the equation:
sec²(θ) * dθ/dt = (-3 * 5 - 7 * 2) / (5²) = (-15 - 14) / 25 = -29 / 25
Now, we can solve for dθ/dt, the rate at which the angle is changing:
dθ/dt = (-29 / 25) * (1 / sec²(θ))
Since sec²(θ) is the reciprocal of cos²(θ), we can rewrite this as:
dθ/dt = (-29 / 25) * (1 / (1 + tan²(θ)))
Now, we substitute the value of tan(θ) that we found previously:
dθ/dt = (-29 / 25) * (1 / (1 + (7 / 5)²))
Simplifying further:
dθ/dt = (-29 / 25) * (1 / (1 + 49 / 25))
= (-29 / 25) * (1 / (74 / 25))
= -29 / 74
Therefore, the rate at which the angle between the x-axis and the line joining the two particles is changing at that instant is -29/74 units per second.
To determine if the particles are moving towards or away from each other at that instant, we need to check the relative velocity vector between the particles.
The relative velocity vector is given by subtracting the velocity vector of particle B from the velocity vector of particle A. The velocity vector of A is (2,0) and the velocity vector of B is (0,-3).
Relative velocity vector = Velocity vector of A - Velocity vector of B
= (2,0) - (0,-3)
= (2,3)
Therefore, the particles are moving towards each other at that instant, as the relative velocity vector has both x and y components positive.