A particle moves along the parabola y=x^2 in the first quadrant and its x-coordinate increases at a steady 10m/s. How fast is the angle of inclination of the line joining the particle to the origin...

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taylormath's profile pic

Posted on

As the angle inclination changes, so will the x and y coordinates.  Three rates need to be considered to answer this question, d(theta)/dt, dx/dt and dy/dt.  The only one given is dx/dt which is 10 m/s.  Since the particle travels the path y=x^2, dy/dt = 2x*dx/dt by taking the derivative with respect to t.

dy/dt = 2x*dx/dt = 2(3)(10) = 60.

To find d(theta)/dt:

A triangle is drawn using the line that makes the angle of inclination.  At x=3, the y-value is 9, since y = x^2 = 3^2 = 9.

Theta is the angle formed by the x-axis and the line that goes through (3,9).

Equation with theta:



Take the derivative of both sides with respect to t:

`(d theta)/dt=1/(1+(y/x)^2)*[(x*(dy/dt)-y*(dx/dt))/x^2]`

`(d theta)/dt=1/(1+(9/3)^2)*[(3(60)-9(10))/3^2]`

`(d theta)/dt=1/10*[(180-90)/9]`

`(d theta)/dt=1/10*10=1`

The angle of inclination is changing at a rate of 1 radian/sec.


thilina-g's profile pic

Posted on

Let angle of inclination be denoted by A, then tan A = y/x, and A = tan^-1(y/x)

We have to calculate the derivative, dA/dt.

dx/dt is given as 3 ms^-1.

dA/dt = d(tan^-1(y/x))/dt. Let Z = y/x


dA/dt = [1 / (1+Z^2)] * dZ/dt . This equation B

Let us find dZ/dt,

Z = y/x therefore,

dZ/dt = [(dy/dt *x) - (1*y)]/x^2

dZ/dt = (x*dy/dt - y)/ x^2. This is equation C


Now we will calculate dy/dt at x = 3m.

y = x^2, y = 3^2 = 9 at x =3m.


Then dy/dt = 2*x*dx/dt

which gives, dy/dt = 2*3*10 =60 at x = 3m.

therefore substituting in equation C gives,

dZ/dt = (3*60-9)/9 = (20-1) = 19.

and Z = y/x = 9/3 = 3.

Now substituting in equation B gives,


dA/dt =[(1 / (1+3^2)] * 19

dA/dt = (1/10)*19 = 1.9 rad/s


Therefore the angle A is changing at a rate of 1.9 rad/s when x = 3m.


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