Asked by Becca
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!
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!
Answers
Answered by
Reiny
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 <b>decreasing</b> at 29/370 radians/second
check my arithmetic
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 <b>decreasing</b> at 29/370 radians/second
check my arithmetic
Answered by
Steve
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
------------------------------
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
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
Answered by
Reiny
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
There are no AI answers yet. The ability to request AI answers is coming soon!