Question

Particle Motion A particle \(P(x, y)\) is moving in the co- ordinate plane in such a way that \(d x / d t=-1 \mathrm{m} / \mathrm{sec}\) and \(d y / d t=-5 \mathrm{m} / \mathrm{sec} .\) How fast is the particle's distance from the origin changing as it passes through the point \((5,12) ?\)

Short answer

The rate of change of the particle's distance from the origin when it passes through the point \((5,12)\) is \(dd/dt\) m/sec.

Step-by-step solution

Formulate the Expression for Distance from the Origin

Using the Pythagorean theorem, we know that the distance \(d\) from the origin can be expressed as \(d = \sqrt{x^2 + y^2}\).

Differentiate the Distance with Respect to Time

To find how the distance changes with time, we need to differentiate \(d\) with respect to \(t\). Use the chain rule. The derivative \(dd/dt\) yields \(dd/dt = (x(dx/dt) + y(dy/dt))/d\). Here, \(dx/dt = -1\) m/sec and \(dy/dt = -5\) m/sec are given. \(x\) and \(y\) will be the coordinates of the point of interest, which is \((5,12)\).

Substitute the Values into dd/dt

Substitute \(dx/dt = -1\) m/sec, \(dy/dt = -5\) m/sec, \(x = 5\), \(y = 12\), and \(d = \sqrt{5^2 + 12^2}\) into \((x(dx/dt) + y(dy/dt))/d\), simplifying to find \(dd/dt\).

Calculate the Result

Perform the necessary computations to obtain the final result for \(dd/dt\) in m/sec.

Key concepts

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry that relates the three sides of a right-angled triangle. It states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This can be written as an equation: \( c^2 = a^2 + b^2 \) where \( c \) represents the length of the hypotenuse, and \( a \) and \( b \) represent the lengths of the triangle's other two sides.

In the context of particle motion, we often use this theorem to determine the distance of a moving particle from the origin in a two-dimensional coordinate system. Here, the particle's position coordinates, \( x \) and \( y \) form the perpendicular sides \( a \) and \( b \) of the right triangle, whereas the distance \( d \) from the origin to the particle makes up the hypotenuse \( c \) of the triangle. Thus, the particle's distance \( d \) is given by \( d = \sqrt{x^2 + y^2} \) which is directly derived from the Pythagorean theorem. Knowing this relationship is crucial when we want to understand how this distance changes over time, which introduces differentiation to measure the rate of change.

Differentiation

Differentiation is a key operation in calculus that measures how a function changes as its input changes. It answers questions about the rate at which quantities vary. The derivative of a function at a particular point gives the slope of the tangent line to the graph of the function at that point, which indicates how fast the function is changing at that moment.

To find the rate of change of the particle's distance from the origin with respect to time, we differentiate the distance function \( d(t) = \sqrt{x(t)^2 + y(t)^2} \) with respect to time \( t \). This involves the application of the chain rule and often requires derivatives of other functions, known as the function's components, which in our problem are the rates of change of the particle's \( x \) and \( y \) coordinates. Differentiating these components separately as \( dx/dt \) and \( dy/dt \) with given values, and then substituting them into the differentiated distance function will yield the rate at which the particle's distance from the origin is changing over time.

Chain Rule

The chain rule is a powerful differentiation rule in calculus for computing the derivative of a composite function. It states that if two functions are composed, then the derivative of the composition is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

For example, if a particle's position is given by a function \( f(t) \) and its distance from the origin is \( d(f(t)) \) where \( f(t) \) represents the inner function and \( d \) represents the outer function, then the chain rule tells us how to find the derivative of \( d \) with respect to \( t \) which is \( \frac{dd}{dt} = \frac{dd}{df} \times \frac{df}{dt} \).

In our exercise, this equates to finding \( \frac{dd}{dt} \) by differentiating the distance \( d = \sqrt{x^2 + y^2} \) with respect to \( t \). The differentiation of the square root is made simpler by using the chain rule, which enables us to first treat \( x^2 + y^2 \) as a single function to be differentiated and then to multiply by the derivative of \( x^2 \) and \( y^2 \) with respect to \( t \) separately. The given rates \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) are used in this context to find the overall rate of change of the distance \( d \) at the specific point \( (5,12) \) as the particle moves.