Question

Compute \(\mathbf{r}^{\prime \prime}(t)\) and \(\mathbf{r}^{\prime \prime \prime}(t)\) for the following functions. $$\mathbf{r}(t)=\langle\cos 3 t, \sin 4 t, \cos 6 t\rangle$$

Short answer

Question: Find the second derivative \(\mathbf{r}^{\prime \prime}(t)\) and the third derivative \(\mathbf{r}^{\prime \prime \prime}(t)\) of the given vector function \(\mathbf{r}(t) = \langle\cos 3t, \sin 4t, \cos 6t\rangle\). Answer: The second derivative \(\mathbf{r}^{\prime \prime}(t)\) is \(\left\langle -9\cos(3t), -16\sin(4t), -36\cos(6t) \right\rangle\), and the third derivative \(\mathbf{r}^{\prime \prime \prime}(t)\) is \(\left\langle 27\sin(3t), -64\cos(4t), 216\sin(6t) \right\rangle\).

Step-by-step solution

Identify the given vector function

The given vector function is $$\mathbf{r}(t) = \langle\cos 3t, \sin 4t, \cos 6t\rangle$$

Find the first derivative \(\mathbf{r}'(t)\)

Differentiate each component of the vector function with respect to \(t\) to obtain the first derivative: $$\mathbf{r}'(t) = \left\langle-\sin(3t)\cdot 3, \cos(4t) \cdot 4, -\sin(6t) \cdot 6\right\rangle$$ Simplify: $$\mathbf{r}'(t) = \left\langle -3\sin(3t), 4\cos(4t), -6\sin(6t) \right\rangle$$

Find the second derivative \(\mathbf{r}''(t)\)

Differentiate each component of the first derivative with respect to \(t\) to obtain the second derivative: $$\mathbf{r}''(t) = \left\langle -3\cos(3t)\cdot 3, -4\sin(4t) \cdot 4, -6\cos(6t) \cdot 6\right\rangle$$ Simplify: $$\mathbf{r}''(t) = \left\langle -9\cos(3t), -16\sin(4t), -36\cos(6t) \right\rangle$$

Find the third derivative \(\mathbf{r}^{\prime \prime \prime}(t)\)

Differentiate each component of the second derivative with respect to \(t\) to obtain the third derivative: $$\mathbf{r}^{\prime \prime \prime}(t) = \left\langle 9\sin(3t)\cdot3, -16\cos(4t) \cdot 4, 36\sin(6t) \cdot 6\right\rangle$$ Simplify: $$\mathbf{r}^{\prime \prime \prime}(t) = \left\langle 27\sin(3t), -64\cos(4t), 216\sin(6t) \right\rangle$$ The second derivative \(\mathbf{r}^{\prime \prime}(t)\) is: $$\mathbf{r}^{\prime \prime}(t) = \left\langle -9\cos(3t), -16\sin(4t), -36\cos(6t) \right\rangle$$ The third derivative \(\mathbf{r}^{\prime \prime \prime}(t)\) is: $$\mathbf{r}^{\prime \prime \prime}(t) = \left\langle 27\sin(3t), -64\cos(4t), 216\sin(6t) \right\rangle$$

Key concepts

Vector Function Differentiation

Differentiating vector functions involves applying the derivative operation to each of the individual components of the vector. When you have a vector function like \( \mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle \), where each component is a function of the variable \( t \), finding the derivative means you'll perform differentiation on each \( f(t) \), \( g(t) \), and \( h(t) \) separately.

In the given problem, the vector function \( \mathbf{r}(t) \) has trigonometric components. Differentiation of these is straightforward, using basic trigonometric derivative rules. For example, the derivative of \( \cos(u) \) regarding a variable \( u \) is \( -\sin(u) \), multiplied by the derivative of \( u \). Thus, if \( u \) is a function of \( t \), like \( 3t \), you apply the chain rule to get the derivative as \( -3\sin(3t) \). This process is used in each component of our vector function, resulting in the first derivative \( \mathbf{r}'(t) \).

Understanding vector function differentiation is vital for solving problems in physics and engineering, where it's often necessary to determine the rate of change of vector quantities.

Second Derivative of Vector Function

The second derivative of a vector function is found by differentiating the first derivative once more. Continuing from the process of finding the first derivative, the second derivative tells us about the acceleration of a particle if we consider our vector function as the position of a particle over time.

Mathematically, to find the second derivative \( \mathbf{r}''(t) \) of the vector function \( \mathbf{r}(t) \), we differentiate each component of \( \mathbf{r}'(t) \) with respect to \( t \). It requires us to use derivative rules once again, which in the case of trigonometric functions, results in switching between cosine and sine functions while also considering any chain rule applications.

For the given exercise, we apply these principles to obtain \( -9\cos(3t) \), \( -16\sin(4t) \), and \( -36\cos(6t) \) as the components of the second derivative vector. This step is crucial for understanding how the direction and magnitude of the vector function's rate of change are also changing.

Third Derivative of Vector Function

Going a step further, the third derivative of a vector function represents the rate of change of the acceleration, which in physics, is sometimes referred to as 'jerk'. To calculate the third derivative \( \mathbf{r}'''(t) \), we differentiate the second derivative components with respect to time \( t \).

The differentiation process is again applied to each component, which involves the standard derivative of trigonometric functions. Using the second derivative obtained before, we compute the third derivatives of the components to get \( 27\sin(3t) \), \( -64\cos(4t) \), and \( 216\sin(6t) \). These values for the third derivative indicate that the vector function’s acceleration itself is a dynamic quantity that can vary over time.

Trigonometric Functions Derivatives

Understanding the derivatives of trigonometric functions is foundational for differentiating vector functions with trigonometric components. The derivatives of sine and cosine, the core trigonometric functions, are straightforward: The derivative of \( \sin(x) \) is \( \cos(x) \) and the derivative of \( \cos(x) \) is \( -\sin(x) \).

Applying these rules, along with the chain rule when the trigonometric functions are composed with another function (like a multiple of \( t \) in our vector function), enables us to handle the differentiation of more complex vector functions that include trigonometric expressions seamlessly. It's important to keep track of signs and multiple angles while differentiating and simplifying expressions.

The use of these derivatives in vector calculus extends to multiple applications, including but not limited to circular motion analysis, wave dynamics, and alternating current in electrical engineering.