Question

Compute the following derivatives. $$\frac{d}{dt}(t^2(\mathbf{i}+2\mathbf{j}-2t\mathbf{k})\cdot (e^t\mathbf{i}+2e^t\mathbf{j}-3e^{-t}\mathbf{k}))$$

Short answer

Question: Determine the time derivative of the dot product of $t^2(\mathbf{i}+2\mathbf{j}-2t\mathbf{k})$ and $(e^t\mathbf{i}+2e^t\mathbf{j}-3e^{-t}\mathbf{k})$. Answer: The time derivative of the dot product is given by: $t^2{e}^t(6e^{-2t}-12t{e}^{-2t})+(5+6t{e}^{-2t})(2t{e}^t+t^2{e}^t)$.

Step-by-step solution

Compute the dot product of the vectors

Computing the dot product of the two given vectors can be done by multiplying their respective components and summing the results: $$t^2(\mathbf{i}+2\mathbf{j}-2t\mathbf{k})\cdot (e^t\mathbf{i}+2e^t\mathbf{j}-3e^{-t}\mathbf{k})=t^2{e}^t(1+2(2)-2t(-3e^{-2t}))$$

Simplify the expression

Now, we can simplify the expression by factoring and combining the terms: $$t^2{e}^t(1+4+6t{e}^{-2t})=t^2{e}^t(5+6t{e}^{-2t})$$

Differentiate the expression

Now, we will differentiate the expression with respect to $t$. We will need to use the product rule and the chain rule: $$\frac{d}{dt}(t^2{e}^t(5+6t{e}^{-2t}))=t^2{e}^t\frac{d}{dt}(5+6t{e}^{-2t})+(5+6t{e}^{-2t})\frac{d}{dt}(t^2{e}^t)$$

Differentiate the inner terms

Now, differentiate the inner terms using the chain rule: $$\frac{d}{dt}(5+6t{e}^{-2t})=0+6e^{-2t}-12t{e}^{-2t}$$ $$\frac{d}{dt}(t^2{e}^t)=2t{e}^t+t^2{e}^t$$

Substitute and simplify

Now, substitute the expressions we found in Step 4 back into the expression from Step 3, and simplify: $$t^2{e}^t(6e^{-2t}-12t{e}^{-2t})+(5+6t{e}^{-2t})(2t{e}^t+t^2{e}^t)$$ The final expression for the derivative is: $$\frac{d}{dt}(t^2(\mathbf{i}+2\mathbf{j}-2t\mathbf{k})\cdot (e^t\mathbf{i}+2e^t\mathbf{j}-3e^{-t}\mathbf{k}))=t^2{e}^t(6e^{-2t}-12t{e}^{-2t})+(5+6t{e}^{-2t})(2t{e}^t+t^2{e}^t)$$

Key concepts

Dot Product

The dot product is a fundamental operation in vector calculus, especially when dealing with vector-valued functions. It allows us to find the scalar result from multiplying two vectors together, which is used in various physics and engineering applications, such as computing work done by a force.

When we take the dot product of two vectors, we multiply their corresponding components and add those products together. For vectors \textbf{A} and \textbf{B} in three-dimensional space with components \textbf{A} = (A_x, A_y, A_z) and \textbf{B} = (B_x, B_y, B_z), their dot product is given by: $\mathbf{\text{A}}\mathbf{\text{B}}=A_x{B}_x+A_y{B}_y+A_z{B}_z$. This operation results in a scalar, not a vector.

In our example, computing the dot product is critical before moving on to differentiation because we want to differentiate a scalar function, not a vector.

Product Rule

The product rule is an essential differentiation technique used when taking the derivative of the product of two functions. It's based on the concept that each function in the product has an 'effect' on the growth rate of the product overall.

The product rule states that if you have two differentiable functions, $u(t)$ and $v(t)$, then the derivative of their product with respect to $t$ is $\frac{d}{dt}(u(t)v(t))=u^{\prime }(t)v(t)+u(t)v^{\prime }(t)$.

In our exercise, the product of $t^2$ and $e^t(5+6t{e}^{-2t})$ must be differentiated using this rule. This step is crucial because it sets the stage for effectively combining and simplifying the resulting derivatives.

Chain Rule

The chain rule is another indispensable tool in calculus. It is used for differentiating composite functions, which are functions comprised of other functions. Essentially, it helps to 'unpack' these nested functions during the differentiation process.

Formally, for two functions $u(t)$ and $v(u)$, where $v$ is a function of $u$ and $u$ is a function of $t$ such as $v(u(t))$ the chain rule tells us that $\frac{dv}{dt}=\frac{dv}{du}\frac{du}{dt}$.

In the exercise, we applied the chain rule to differentiate the term $6t{e}^{-2t}$, which is a product involving an exponential function of $t$—thus, a composition of functions. This is critical for obtaining the correct derivative of the expression and correctly simplifying the result.

Differentiation of Exponential Functions

Exponential functions, often represented as $e^x$, have a unique property where the derivative is proportional to the function itself. Specifically, the derivative of $e^x$ with respect to $x$ is $e^x$. However, when the exponent is a function of $x$ itself, such as $e^{f(x)}$, we need to apply the chain rule to differentiate it.

In our context, $e^t$ and $e^{-2t}$ are both exponential functions, yet they involve the variable $t$, requiring the chain rule for proper differentiation. Correctly applying the differentiation rules to exponential functions ensures that the final derivative accounts for all aspects of their growth rates—which can be especially rapid due to the nature of exponentials.