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
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!
3 answers
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
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
------------------------------
second method
------------------------------
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
I must have made an error in my arithmetic, too lazy to find it
1 answer