Problem 41
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?
Step-by-Step Solution
Verified Answer
The velocity vector makes a 45° angle with the axes at time \( t = \frac{2b}{3c} \).
1Step 1: 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:\[ \vec{v} = \frac{d}{dt}(bt^2\hat{\imath} + ct^3\hat{\jmath}) = 2bt\hat{\imath} + 3ct^2\hat{\jmath} \]
2Step 2: 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 \( \tan(45^\circ) = 1 \).
3Step 3: Set Up Equation
From the condition \( \tan(\theta) = 1 \), we have:\[ \frac{3ct^2}{2bt} = 1 \]
4Step 4: 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 functionsangle between vectorsposition and velocity vectors
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.
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.
Other exercises in this chapter
Problem 39
A rocket is fired at an angle from the top of a tower of height \(h_0\) = 50.0 m. Because of the design of the engines, its position coordinates are of the form
View solution Problem 40
A faulty model rocket moves in the \(xy\)-plane (the positive \(y\)-direction is vertically upward). The rocket's acceleration has components \(a_x(t) = \alpha
View solution Problem 42
The position of a dragonfly that is flying parallel to the ground is given as a function of time by \(\vec{r} = [2.90 m + (0.0900 m/s^2)t^2] \hat{\imath} - (0.0
View solution Problem 44
A bird flies in the \(xy\)-plane with a velocity vector given by \(\overrightarrow{v}\) = \((a - \beta$$t^{2})\) \(\hat{l}+\gamma t \hat{j}\), with \(\alpha = 2
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