# A particle is constrained to move in #Q_1#, in a parabolic groove # y = x^2#. It was at O, at t = 0 s. At time t s, its speeds in the directions Ox and Oy are 5 and 12 cm/s, respectively. How do you prove its direction makes #67^o22'49''# with Ox?

See below.

Because of the constraint, the velocity vector is also the tangent vector. Therefore:

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To prove that the direction of the particle makes an angle of 67°22'49'' with Ox, you can use trigonometry. The tangent of the angle between the direction of motion and the x-axis is given by the ratio of the velocity component in the y-direction to the velocity component in the x-direction.

Let ( v_x ) be the velocity component in the x-direction and ( v_y ) be the velocity component in the y-direction.

The tangent of the angle ( \theta ) between the direction of motion and the x-axis is given by:

[ \tan(\theta) = \frac{v_y}{v_x} ]

Substitute the given values:

[ \tan(\theta) = \frac{12}{5} ]

Now, find the angle ( \theta ) using the inverse tangent function (arctan):

[ \theta = \arctan\left(\frac{12}{5}\right) ]

Calculate ( \theta ) in degrees, minutes, and seconds.

[ \theta = 67°22'49'' ]

Therefore, the direction of the particle makes an angle of ( 67°22'49'' ) with the x-axis (Ox).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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