Question

A particle is moving along the curve y= 5 \sqrt{4 x + 5}. As the particle passes through the point (1, 15), its x-coordinate increases at a rate of 2 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

Answer 1

In order to find the rate of change of the distance from the particle to the origin, we can use the Pythagorean theorem. Let's call the distance from the particle to the origin at a given point (x, y) as d.
Using the Pythagorean theorem, we have:
\( d^2 = x^2 + y^2 \)
Taking the derivative with respect to t (time), we have:
\( 2d \frac{dd}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \)
Noticed that we are given that the x-coordinate increases at a rate of 2 units per second, so \( \frac{dx}{dt} = 2 \) units per second.
Furthermore, at the point (1, 15), we are also given that \( x = 1 \) and \( y = 15 \).
Substituting these values into our derivative equation:
\( 2d \frac{dd}{dt} = 2 \cdot 1 \cdot 2 + 2 \cdot 15 \cdot \frac{dy}{dt} \)
\( 2d \frac{dd}{dt} = 4 + 30 \frac{dy}{dt} \)
Since in the expression \( y= 5 \sqrt{4 x + 5} \), x and y are related, we can substitute the expression for y in the derived equation in the given expression:
\( (5^2) (4 x + 5) = x^2 + y^2 \)
\( 20x + 25 = x^2 + y^2 \)
We can then substitute this relationship of x and y regarding the expression for \( \frac{dy}{dt} \):
\( 2d \frac{dd}{dt} = 4 + 30 \cdot \frac{-5x}{20x + 25} \)
At (1, 15), this gives \( 2d \frac{dd}{dt} = 4 + 30 \cdot \frac{-5}{20 + 25} \)
\( 2d \frac{dd}{dt} = 4 - 30 \cdot \frac{5}{45} \)
\( \frac{dd}{dt} = \frac{-4}{2d} + \frac{-3}{3} \)
\( \frac{dd}{dt} = -\frac{7}{3} \)
Therefore, the rate of change of the distance from the particle to the origin at this instant is \( -\frac{7}{3} \) units per second.