Question

Find the function \(\mathbf{r}\) that satisfies the given conditions. $$\mathbf{r}^{\prime}(t)=\left\langle e^{t}, \sin t, \sec ^{2} t\right\rangle ; \mathbf{r}(0)=\langle 2,2,2\rangle$$

Short answer

Question: Determine the vector-valued function \(\mathbf{r}(t)\) that satisfies the given conditions: \(\mathbf{r}'(t) = \langle e^t, \sin t, \sec^2 t \rangle\) and \(\mathbf{r}(0) = \langle 2,2,2 \rangle\). Answer: The vector-valued function \(\mathbf{r}(t)\) satisfying the given conditions is \(\mathbf{r}(t) = \left\langle e^{t} + 1, -\cos t - 1, \tan t + 2 \right\rangle\).

Step-by-step solution

Integrate each component of the vector-valued function

To find \(\mathbf{r}(t)\), integrate each component of \(\mathbf{r}'(t)\) with respect to \(t\). $$\int e^{t} \, dt = e^{t} + C_1$$ $$\int \sin t \, dt = -\cos t + C_2$$ $$\int \sec^2 t \, dt = \tan t + C_3$$ So, we have: $$\mathbf{r}(t) = \left\langle e^{t} + C_1, -\cos t + C_2, \tan t + C_3 \right\rangle$$

Apply the initial condition

We are given \(\mathbf{r}(0) = \langle 2, 2, 2 \rangle\). Plug \(t=0\) into the expression of \(\mathbf{r}(t)\) and set it equal to the initial condition. $$\mathbf{r}(0) = \left\langle e^{0} + C_1, -\cos 0 + C_2, \tan 0 + C_3 \right\rangle = \langle 1 + C_1, 1 - C_2, C_3 \rangle$$ Now, set the components equal: $$1 + C_1 = 2 \implies C_1 = 1$$ $$1 - C_2 = 2 \implies C_2 = -1$$ $$C_3 = 2$$

Plug in the constants to find \(\mathbf{r}(t)\)

Substitute the values of the constants \(C_1, C_2,\) and \(C_3\) back into the expression for \(\mathbf{r}(t)\) to obtain the final solution: $$\mathbf{r}(t) = \left\langle e^{t} + 1, -\cos t - 1, \tan t + 2 \right\rangle$$ So, the vector-valued function \(\mathbf{r}(t)\) that satisfies the given conditions is: $$\mathbf{r}(t) = \left\langle e^{t} + 1, -\cos t - 1, \tan t + 2 \right\rangle$$

Key concepts

Integration of Vector-Valued Functions

Understanding how to integrate vector-valued functions is a cornerstone in calculus, especially when analyzing the motion and paths in physics and engineering. A vector-valued function is essentially a function that has vectors as its output instead of scalars. To integrate this kind of a function, you perform the integration process on each of the individual scalar components of the vector.

Imagine each component as a separate function tied to an axis in a coordinate system. For instance, if your vector-valued function is \( \mathbf{r}'(t) = \< e^t, \sin t, \sec^2 t \> \) as in the given exercise, you integrate each part of \( \mathbf{r}'(t) \) with respect to \( t \) individually, just like scalar functions.

The integration of \( e^t \) gives us \( e^t \) due to the exponential rule, \( \sin t \) integrates to \( -\cos t \) following trigonometric integration rules, and \( \sec^2 t \) integrates to \( \tan t \) by the direct antiderivative.

After integration, each component often includes an arbitrary constant of integration \( (C_1, C_2, C_3, ...) \) since the antiderivative of a function is not unique. Combining these gives us a new vector-valued function, \( \mathbf{r}(t) \), representing our integrated result which will include these constants. If initial conditions are available, they can be applied to determine these constants.

Initial Conditions in Calculus

Initial conditions in calculus are essential to determine the specific solution to a differential equation when the general solution involves arbitrary constants. These conditions represent the state of the function at a particular time or position and serve as constraints to find a unique solution.

In the context of our exercise, the initial condition \( \mathbf{r}(0) = \<2,2,2\> \) provides the specific value of the vector-valued function at time \( t = 0 \). Using this information, we can solve for the constants \( C_1, C_2, \) and \( C_3 \) in our integrated function \( \mathbf{r}(t) \).

By substituting \( t = 0 \) into the general solution, each component of the initial condition coincides with its respective component in the general solution (after evaluating at \( t = 0 \) and considering the constants), so equating the corresponding parts we derive the values for these constants. This step transforms our general solution into a particular solution that uniquely satisfies the given condition.

Applying initial conditions correctly is crucial in many areas, such as solving motion problems in physics or determining growth patterns in biology.

Trigonometric Integration

Trigonometric integration refers to the process of finding the antiderivative of a trigonometric function, a common task in calculus that is related to solving for the area under a curve or determining the displacement for a given velocity function.

To integrate basic trigonometric functions, one must be familiar with their standard antiderivatives, such as \( \int \sin x \, dx = -\cos x + C \) or \( \int \sec^2 x \, dx = \tan x + C \). However, trigonometric integration may also involve more complex scenarios, such as when using identities to simplify expressions before integration.

For example, knowing that \( \sec^2 x \) is the derivative of \( \tan x \) allows for direct integration, but what if the integral is not so straightforward? In such cases, trigonometric identities like the Pythagorean identities, angle sum and difference formulas, and double-angle formulas become incredibly useful. The more you practice these techniques, the more intuitive they will become, making it easier to tackle varied and complex trigonometric integrals.