Question

I am looking for assistance in starting to solve this problem. Dont need the answer but a methodology for figuring it out on my own. A particle moves from right to left along the parabolic curve y = square root of -x in such a way that its x coordinates decreases at the rate of 4 meters per second. How fast is the angle of inclination in degrees of the line joining the particle to the origin changing when x = -2?

Answer 1

Problem:

I am looking for assistance in starting to solve this problem. Don't need the answer but a methodology for figuring it out on my own. A particle moves from right to left along the parabolic curve $y =$ square root of $-x$ in such a way that its $x$ coordinates decreases at the rate of 4 meters per second. How fast is the angle of inclination in degrees of the line joining the particle to the origin changing when $x = -2$?

Solution:

According to the problem statement:

$$y = \sqrt {- x}$$

And

$$v = \frac {d x}{d t} = - 4 \, \mathrm{m/s}$$

After differentiation of the first equation:

$$\frac {d y}{d x} = t g \alpha = - \frac {1}{2 \sqrt {- x}}$$

Where $\alpha$ – angle of inclination.

Differentiation of the equation (3) with respect to time $t$ gives:

$$\frac {d (t g \alpha)}{d t} = \frac {1}{\cos^ {2} \alpha} \frac {d \alpha}{d t} = - \frac {1}{4} (- x) ^ {- \frac {3}{2}} \frac {d x}{d t} = >$$ $$\frac {d \alpha}{d t} = \frac {\cos^ {2} \alpha}{4} (- x) ^ {- \frac {3}{2}} v$$

From equation (3) we get:

$$\cos^ {2} \alpha = \frac {1}{1 + t g ^ {2} \alpha} = \frac {1}{1 + \frac {1}{- 4 x}} = \frac {4 x}{4 x - 1}$$

Then equation (4):

$$\frac {d \alpha}{d t} = \frac {x}{(4 x - 1)} (- x) ^ {- \frac {3}{2}} v$$

Since $x = -2$ :

$$\frac {d \alpha}{d t} = \frac {- 2}{(4 * (- 2) - 1)} (2) ^ {- \frac {3}{2}} * (- 4) = 0.31 \frac {rad}{s} = 18 \frac {degrees}{s}$$

Answer: $\frac{d\alpha}{dt} = 18\frac{degrees}{s}$ (at point $x = -2$).