Points A and B move along the x and y axes, respectively, in such a way that the distance r (meters) along the perpendicular from the origin to the line AB remains constant. How fast is OA changing, and is it increasing or decreasing, when OB = 2r and B is moving toward O at the rate of 0.3r m/sec?
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Related Rates Problem
- Thread starter dsfjdkfj
- Start date
You could let d be the distance from the origin to the point on the line.
A would have coordinates (a,0) and B coordinates (0,b)
Then, b would have length \(\displaystyle b=Dsec\theta\) and \(\displaystyle a=D\cdot csc\theta\)
So, by Pythagoras, \(\displaystyle (D/b)^{2}+(D/a)^{2}=1\)
\(\displaystyle \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{D^{2}}\)
Now, differentiate w.r.t t and get \(\displaystyle \frac{2}{a^{3}}\frac{da}{dt}+\frac{2}{b^{3}}\frac{db}{dt}=0\). Since D remains constant, dD/dt=0 on the right side.
Now, can you continue using the given info?.
A would have coordinates (a,0) and B coordinates (0,b)
Then, b would have length \(\displaystyle b=Dsec\theta\) and \(\displaystyle a=D\cdot csc\theta\)
So, by Pythagoras, \(\displaystyle (D/b)^{2}+(D/a)^{2}=1\)
\(\displaystyle \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{D^{2}}\)
Now, differentiate w.r.t t and get \(\displaystyle \frac{2}{a^{3}}\frac{da}{dt}+\frac{2}{b^{3}}\frac{db}{dt}=0\). Since D remains constant, dD/dt=0 on the right side.
Now, can you continue using the given info?.
Here is a link I just found. I was thinking along the same lines. This can explain it.
http://www.mth.msu.edu/users/mccarthy/MTH132/problem.118.pdf
http://www.mth.msu.edu/users/mccarthy/MTH132/problem.118.pdf
wow that helped tremendously thanks a lot. however, i still have a few questions.
how do you get from: (− 2)(a^−3 )(da/dt ) + (− 2)(y^−3 )(dy/dt ) = 0 to y^3( da/dt) + a^3( dy/dt) = 0 ?
also, (8r^3 )(da/dt ) + (8r^3/ 33/2 )(− 0. 3r ) = 0
solve for da/dt:
da/dt = 0 .3 /3^3/2 = 0 .1 /3^1/2 = 3^1/2 /30
where did the r go in this equation?
i eventually got 0.3r/3^3/2 ad da/dt
how do you get from: (− 2)(a^−3 )(da/dt ) + (− 2)(y^−3 )(dy/dt ) = 0 to y^3( da/dt) + a^3( dy/dt) = 0 ?
also, (8r^3 )(da/dt ) + (8r^3/ 33/2 )(− 0. 3r ) = 0
solve for da/dt:
da/dt = 0 .3 /3^3/2 = 0 .1 /3^1/2 = 3^1/2 /30
where did the r go in this equation?
i eventually got 0.3r/3^3/2 ad da/dt
D
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.wow that helped tremendously thanks a lot. however, i still have a few questions.
how do you get from: (− 2)(a^−3 )(da/dt ) + (− 2)(y^−3 )(dy/dt ) = 0
multiply by [-a^3 * y^3]/2
to y^3( da/dt) + a^3( dy/dt) = 0 ?
also, (8r^3 )(da/dt ) + (8r^3/ 33/2 )(− 0. 3r ) = 0
solve for da/dt:
da/dt = 0 .3 /3^3/2 = 0 .1 /3^1/2 = 3^1/2 /30
where did the r go in this equation?
i eventually got 0.3r/3^3/2 ad da/dt