Question

Compute the following derivatives. $$\frac{d}{d t}\left(\left(3 t^{2} \mathbf{i}+\sqrt{t} \mathbf{j}-2 t^{-1} \mathbf{k}\right) \cdot(\cos t \mathbf{i}+\sin 2 t \mathbf{j}-3 t \mathbf{k})\right)$$

Short answer

Question: Compute the time derivative of the dot product of the vector functions \(3t^2 \mathbf{i}+\sqrt{t}\mathbf{j}-2t^{-1}\mathbf{k}\) and \(\cos t\mathbf{i}+\sin 2t\mathbf{j}-3t\mathbf{k}\). Answer: The time derivative of the dot product of the given vector functions is \(-6t\sin{t}+6\cos{t} +(2\cos{2t}-3)t^{-\frac{1}{2}}-\frac{6}{t^2}\).

Step-by-step solution

Compute the Derivative of the First Vector Function

To compute the derivative of the first vector function, we need to derive each component with respect to \(t\): 1. Derivative of \(3t^2\) with respect to \(t\): $$\frac{d}{d t}\left(3t^2\right)=6t$$ 2. Derivative of \(\sqrt{t} = t^{1/2}\) with respect to \(t\): $$\frac{d}{d t}\left(t^{1/2}\right)=\frac{1}{2}t^{-1/2}$$ 3. Derivative of \(-2t^{-1}=-2/t\) with respect to \(t\): $$\frac{d}{d t}\left(-2t^{-1}\right)=2t^{-2}$$ Thus, the derivative of the first vector function is: $$\frac{d}{dt}\left(3t^2\mathbf{i} + \sqrt{t}\mathbf{j} -2 t^{-1}\mathbf{k}\right)=6t\mathbf{i} +\frac{1}{2}t^{-1/2}\mathbf{j} + 2t^{-2}\mathbf{k}$$

Compute the Derivative of the Second Vector Function

To compute the derivative of the second vector function, we need to derive each component with respect to \(t\): 1. Derivative of \(\cos t\) with respect to \(t\): $$\frac{d}{dt}\left(\cos t\right)=-\sin t$$ 2. Derivative of \(\sin 2t\) with respect to \(t\): $$\frac{d}{dt}\left(\sin 2t\right)=2\cos 2t$$ 3. Derivative of \(-3t\) with respect to \(t\): $$\frac{d}{dt}\left(-3t\right)=-3$$ Thus, the derivative of the second vector function is: $$\frac{d}{dt}\left(\cos t\mathbf{i} +\sin 2t\mathbf{j} -3t\mathbf{k}\right)=-\sin t\mathbf{i} +2\cos 2t\mathbf{j} -3\mathbf{k}$$

Compute the Dot Product between the First Vector's Derivative and the Second Vector

To compute the dot product, we multiply corresponding components and sum the products: $$\begin{aligned} \left(6t\mathbf{i} +\frac{1}{2}t^{-1/2}\mathbf{j} +2t^{-2}\mathbf{k}\right) \cdot \left(\cos t\mathbf{i} +\sin 2t\mathbf{j} -3t\mathbf{k}\right) &= 6t\cos t +\frac{1}{2}t^{-1/2}\sin 2t- 6t^{-1} \\ &= 6t\cos t +\frac{1}{2}t^{-1/2}\sin 2t- 6/t \end{aligned}$$

Compute the Derivative of the Dot Product

Finally, we compute the derivative of the dot product with respect to t: $$\frac{d}{dt}\left(6t\cos{t} + \frac{1}{2}t^{-\frac{1}{2}}\sin{2t} - \frac{6}{t}\right)=-6t\sin{t}+6\cos{t} + (2\cos{2t}-3)t^{-\frac{1}{2}}-\frac{6}{t^2}$$ Thus, the final answer is: $$\frac{d}{dt}\left(\left(3t^2 \mathbf{i}+\sqrt{t}\mathbf{j}-2t^{-1}\mathbf{k}\right) \cdot(\cos t\mathbf{i}+\sin 2t\mathbf{j}-3t\mathbf{k})\right)=-6t\sin{t}+6\cos{t} +(2\cos{2t}-3)t^{-\frac{1}{2}}-\frac{6}{t^2}$$

Key concepts

Dot Product Differentiation

Understanding dot product differentiation is essential when dealing with vector functions in calculus. The key is to apply the product rule, which is commonly known in scalar functions, to vector functions.

When we differentiate a dot product of two vector functions, each with respect to a common variable, we differentiate one vector and dot it with the second, then add to it the first vector dotted with the derivative of the second vector. This can be expressed mathematically as:
\[\begin{equation}\frac{d}{dt}(\textbf{u} \. \textbf{v}) = \frac{d\textbf{u}}{dt} \. \textbf{v} + \textbf{u} \. \frac{d\textbf{v}}{dt}\end{equation}\]
where \textbf{u} and \textbf{v} are vector functions of the variable t.

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Application in the Given Exercise

The approach is breaking down the problem into steps: first differentiating each vector component-wise, and then applying the dot product to the derivatives as visualized in the solution provided. The differentiation of each individual component relies on standard calculus rules, such as the power rule.

Vector Calculus

Vector calculus is a field within mathematics that deals with vector fields and vector-valued functions. Key operations in vector calculus include vector addition, scalar multiplication, dot product, cross product, and differentiation. Each of these operations has specific rules that help us describe and analyze multidimensional space phenomena, like physical forces or velocity fields.

The calculus of vector functions behaves much like the calculus of scalar functions, but it needs careful handling of each component and the use of vector-specific operations like the dot product, as seen in the provided exercise.

Chain Rule Calculus

The chain rule is a formula for computing the derivative of the composition of two or more functions. In the context of multivariable calculus and vector calculus, the chain rule can be extended to deal with vector functions and their compositions.

If a vector function is the product of scalar and vector functions, you'd apply the chain rule when differentiating this composition. The process involves taking derivatives of the outer and inner functions and then 'chaining' them together. Applying the chain rule correctly is a critical step in finding the derivative of a vector function in relation to a single variable.

As vector functions can have multiple components, applying the chain rule typically requires you to address each component individually, as illustrated in the step-by-step solution of the textbook exercise.

Multivariable Calculus

Multivariable calculus extends single-variable calculus concepts to functions of several variables. It encompasses topics such as partial derivatives, multiple integrals, and vector calculus. It allows us to handle functions that depend on more than one variable and analyze phenomena where several variables interact simultaneously.

In the dot product differentiation problem presented, the principles of multivariable calculus apply because we're dealing with vector functions that inherently involve multiple variables, one for each component of the vector. The step-by-step solution showcases how to deal with such a scenario: by carefully managing each part of the vectors and applying differentiation rules to each component separately.