Chapter 14: Problem 13
Differentiate the following functions. $$\mathbf{r}(t)=e^{t} \mathbf{i}+2 e^{-t} \mathbf{j}-4 e^{2 t} \mathbf{k}$$
Short Answer
Expert verified
Answer: The derivative of the vector function with respect to \(t\) is \(\frac{d\mathbf{r}}{dt} = e^t\mathbf{i} - 2e^{-t}\mathbf{j} - 8e^{2t}\mathbf{k}\).
Step by step solution
01
Identify the components of the vector function
The given vector function \(\mathbf{r}(t)\) has the following components:
- \(x(t) = e^t\)
- \(y(t) = 2e^{-t}\)
- \(z(t) = -4e^{2t}\)
02
Differentiate the x-component
To differentiate \(x(t) = e^t\) with respect to \(t\), we use the fact that the derivative of \(e^t\) is itself. So, the derivative of the x-component is:
$$\frac{dx}{dt} = e^t$$
03
Differentiate the y-component
To differentiate \(y(t) = 2e^{-t}\) with respect to \(t\), we first find the derivative of \(e^{-t}\) which is \(-e^{-t}\). Then, since we have a constant \(2\), we multiply it with the derivative of the exponential function. So, the derivative of the y-component is:
$$\frac{dy}{dt} = -2e^{-t}$$
04
Differentiate the z-component
To differentiate \(z(t) = -4e^{2t}\) with respect to \(t\), we first find the derivative of \(e^{2t}\) which is \(2e^{2t}\). Then, since we have a constant \(-4\), we multiply it with the derivative of the exponential function. So, the derivative of the z-component is:
$$\frac{dz}{dt} = -8e^{2t}$$
05
Write the derivative of the vector function
Now, we can write the derivative of the vector function \(\mathbf{r}(t)\) by combining the derivatives of each component in the form of a new vector function:
$$\frac{d\mathbf{r}}{dt} = \frac{dx}{dt}\mathbf{i} + \frac{dy}{dt}\mathbf{j} + \frac{dz}{dt}\mathbf{k} = e^t\mathbf{i} - 2e^{-t}\mathbf{j} - 8e^{2t}\mathbf{k}$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
vector functions
In calculus, a vector function represents a vector quantity that depends on a single parameter, often time \(t\). This is incredibly useful in physics and engineering for expressing the position, velocity, or acceleration of an object in two- or three-dimensional space.
A vector function is written in the form \(\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k}\), where \(x(t)\), \(y(t)\), and \(z(t)\) are scalar functions of \(t\) corresponding to each component.
Key points about vector functions:
A vector function is written in the form \(\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k}\), where \(x(t)\), \(y(t)\), and \(z(t)\) are scalar functions of \(t\) corresponding to each component.
Key points about vector functions:
- They can describe motion in space.
- Each component operates independently.
- Can be differentiated or integrated like scalar functions, giving valuable insights into physical processes.
derivatives of exponential functions
The derivative of exponential functions is a foundational concept in calculus. Exponential functions with base \(e\), such as \(e^t\) and \(e^{-t}\), are particularly important due to their unique property: their rate of growth is proportional to their current value.
The derivative of \(e^t\) with respect to \(t\) is simply \(e^t\). When you have a function like \(e^{-t}\), its derivative is \(-e^{-t}\), introducing a negative sign due to the chain rule.
For functions with different exponents, like \(e^{kt}\), the derivative involves multiplying by the constant \(k\), leading to \(ke^{kt}\). This shows how the exponent affects the growth or decay rate.
The derivative of \(e^t\) with respect to \(t\) is simply \(e^t\). When you have a function like \(e^{-t}\), its derivative is \(-e^{-t}\), introducing a negative sign due to the chain rule.
For functions with different exponents, like \(e^{kt}\), the derivative involves multiplying by the constant \(k\), leading to \(ke^{kt}\). This shows how the exponent affects the growth or decay rate.
- Exponential growth is when the function rapidly increases.
- Exponential decay is when the function decreases.
- The chain rule helps in differentiating compositions involving exponential functions.
component-wise differentiation
Component-wise differentiation involves taking the derivative of each individual component of a vector function separately. This method simplifies the process of differentiation in multi-dimensional calculus, as each component function can be treated as an independent function, just like scalar functions.
To differentiate a vector function \(\mathbf{r}(t)\), we differentiate each component independently:
To differentiate a vector function \(\mathbf{r}(t)\), we differentiate each component independently:
- \(x(t) = e^t\) results in \(\frac{dx}{dt} = e^t\).
- \(y(t) = 2e^{-t}\) results in \(\frac{dy}{dt} = -2e^{-t}\).
- \(z(t) = -4e^{2t}\) results in \(\frac{dz}{dt} = -8e^{2t}\).