Question

Let \(\mathbf{u}(t)=\left\langle 1, t, t^{2}\right\rangle, \mathbf{v}(t)=\left\langle t^{2},-2 t, 1\right\rangle\) and \(g(t)=2 \sqrt{t}\). Compute the derivatives of the following functions. $$\mathbf{v}\left(e^{t}\right)$$

Short answer

Answer: The derivative of \(\mathbf{v}\left(e^t\right)\) is \(\left\langle 2e^{2t}, -2e^t, 0 \right\rangle\).

Step-by-step solution

Identify the given functions

We are given the following functions: $$\mathbf{u}(t) = \left\langle 1, t, t^2 \right\rangle$$ $$\mathbf{v}(t) = \left\langle t^2,-2t, 1 \right\rangle$$ $$g(t) = 2\sqrt{t}$$ Our goal is to find the derivative of \(\mathbf{v}\left(e^t\right)\).

Apply the chain rule to \(\mathbf{v}\left(e^t\right)\)

To compute the derivative of \(\mathbf{v}\left(e^t\right)\), we will use the chain rule. The chain rule states that if we have a function \(\mathbf{v}(g(t))\), its derivative with respect to \(t\) is given by: $$\frac{d}{dt}\mathbf{v}\left(g(t)\right) = \frac{d\mathbf{v}}{dg} \cdot \frac{dg}{dt}$$ In our case, \(g(t) = e^t\), so we want to find: $$\frac{d}{dt}\mathbf{v}\left(e^t\right) = \frac{d\mathbf{v}}{de^t} \cdot \frac{de^t}{dt}$$

Compute \(\frac{de^t}{dt}\) and \(\frac{d\mathbf{v}}{dt}\)

First, let's find the derivative of \(e^t\) with respect to \(t\): $$\frac{de^t}{dt} = e^t$$ Next, let's compute the derivative of \(\mathbf{v}\) with respect to \(t\): $$\frac{d\mathbf{v}}{dt} = \left\langle \frac{d}{dt}(t^2), \frac{d}{dt}(-2t), \frac{d}{dt}(1) \right\rangle = \left\langle 2t, -2, 0 \right\rangle$$

Compute \(\frac{d\mathbf{v}}{de^t}\)

Now, we substitute \(e^t\) for \(t\) in the derivative of \(\mathbf{v}\) with respect to \(t\): $$\frac{d\mathbf{v}}{de^t} = \left\langle 2e^t, -2, 0 \right\rangle$$

Compute the derivative of \(\mathbf{v}\left(e^t\right)\) using the chain rule

Now we can use the chain rule to compute the derivative of \(\mathbf{v}\left(e^t\right)\): $$\frac{d}{dt}\mathbf{v}\left(e^t\right) = \frac{d\mathbf{v}}{de^t} \cdot \frac{de^t}{dt} = \left\langle 2e^t, -2, 0 \right\rangle \cdot e^t = \left\langle 2e^{2t}, -2e^t, 0 \right\rangle$$ So, the derivative of \(\mathbf{v}\left(e^t\right)\) is: $$\frac{d}{dt}\mathbf{v}\left(e^t\right) = \left\langle 2e^{2t}, -2e^t, 0 \right\rangle$$