Question

If \(\vec{r} = bt^2\hat{\imath} + ct^3\hat{\jmath}\), where \(b\) and \(c\) are positive constants, when does the velocity vector make an angle of 45.0\(^\circ\) with the \(x\)- and \(y\)-axes?

Short answer

The velocity vector makes a 45° angle with the axes at time \( t = \frac{2b}{3c} \).

Step-by-step solution

Represent Velocity Vector

The velocity vector \( \vec{v} \) can be found by taking the derivative of the position vector \( \vec{r} \) with respect to time \( t \). Calculating this gives:\[ \vec{v} = \frac{d}{dt}(bt^2\hat{\imath} + ct^3\hat{\jmath}) = 2bt\hat{\imath} + 3ct^2\hat{\jmath} \]

Define Angle Condition

The angle \( \theta \) that the velocity vector makes with the \( x \)-axis is given by:\[ \tan(\theta) = \frac{v_y}{v_x} = \frac{3ct^2}{2bt} \]Since the velocity vector makes a 45.0\(^\circ\) angle with the axes, \( \theta = 45.0\) degrees, and \( \tan(45^\circ) = 1 \).

Set Up Equation

From the condition \( \tan(\theta) = 1 \), we have:\[ \frac{3ct^2}{2bt} = 1 \]

Solve for t

Simplify the equation:\[ \frac{3ct}{2b} = 1 \]Multiply both sides by \( 2b \):\[ 3ct = 2b \]Solve for \( t \):\[ t = \frac{2b}{3c} \]

Key concepts

derivative of vector functions

Understanding the derivative of vector functions is crucial in vector calculus. When we talk about the derivative of a vector function, like the position vector \( \vec{r} = bt^2\hat{\imath} + ct^3\hat{\jmath} \), we want to find how this vector changes with time. This change is represented by the derivative function, often known as the velocity vector. To derive it, we differentiate each component separately with respect to time \( t \). In our example, we take the derivative of \( bt^2 \hat{\imath} \), giving \( 2bt \hat{\imath} \), and the derivative of \( ct^3 \hat{\jmath} \), resulting in \( 3ct^2 \hat{\jmath} \). Hence, the velocity vector is \( \vec{v} = 2bt\hat{\imath} + 3ct^2\hat{\jmath} \).This tells us how fast and in which direction the position vector is changing at any given point in time. Understanding this concept helps us solve problems related to objects in motion.

angle between vectors

In vector calculus, the angle between vectors is an essential concept. It helps describe how one vector orientation relates to another. When the velocity vector \( \vec{v} = 2bt\hat{\imath} + 3ct^2\hat{\jmath} \) makes a specific angle with the coordinate axes, we can use this relationship to find time \( t \).We often use the tangent function to find angles between vectors and axes. The tangent of the angle \( \theta \) that a vector makes with the \( x \)-axis is given by the ratio of its \( y \)-component to its \( x \)-component, \( \tan(\theta) = \frac{v_y}{v_x} \). If the angle is \( 45^\circ \), the tangent value is 1, simplifying your calculations and allowing for easy comparison of components:

• Convert this angle condition into a mathematical equation by setting the ratio equal to 1.
• This equivalency helps solve for unknowns like time \( t \) when given specific angle conditions.

Knowing how to handle these calculations is key for analyzing movement directions and vectors.

position and velocity vectors

Position and velocity vectors are fundamental concepts in physics and vector calculus. A position vector \( \vec{r} \) describes the position of an object in space as a function of time. For example, the vector \( \vec{r} = bt^2\hat{\imath} + ct^3\hat{\jmath} \) gives the object's position along the \( x \)- and \( y \)-axes as time evolves. Velocity vectors, on the other hand, express how fast the object is moving and its direction. It is derived by differentiating the position vector with respect to time. This yields the velocity vector \( \vec{v} = 2bt\hat{\imath} + 3ct^2\hat{\jmath} \). The components of velocity tell us the rate of change along each axis:

• \( 2bt \hat{\imath} \) indicates the velocity in the \( x \)-direction.
• \( 3ct^2 \hat{\jmath} \) describes the velocity in the \( y \)-direction.

By analyzing these vectors, we can determine how objects change position over time and their respective motion paths. Understanding this relationship is essential for predicting future positions and velocities from a set of initial conditions.