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. $$\mathbf{v}(g(t))$$

Short answer

Answer: The derivative of the function \(\mathbf{v}(g(t))\) is \(\frac{d\mathbf{v}(g(t))}{dt} = \left\langle -\frac{1}{t^{5/2}}, \frac{1}{t^{3/2}}, 0 \right\rangle\).

Step-by-step solution

Apply the Chain Rule

The chain rule states that, if we have a function composed of two functions, say \(h(x) = f(g(x))\), then the derivative of \(h(x)\) with respect to \(x\) is equal to the derivative of \(f\) with respect to \(g(x)\) multiplied by the derivative of \(g(x)\) with respect to \(x\). Mathematically, \(\frac{dh(x)}{dx} = \frac{df(g(x))}{dg(x)} \cdot \frac{dg(x)}{dx}\). We will apply the chain rule to find the derivative of \(\mathbf{v}(g(t))\).

Find the derivative of \(\mathbf{v}(t)\)

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

Find the derivative of \(g(t)\)

Now, let's compute the derivative of \(g(t)\) with respect to \(t\). \(\frac{dg(t)}{dt} = \frac{d(2\sqrt{t})}{dt} = \frac{2}{2\sqrt{t}} = \frac{1}{\sqrt{t}}\)

Compute the derivative of \(\mathbf{v}(g(t))\) using the Chain Rule

Now we have all the components needed to apply the chain rule. We will replace \(t\) in the expression \(\mathbf{v}(t)\) with \(g(t)\), then compute the derivative with respect to \(t\), which will be equal to the derivative of \(\mathbf{v}(g(t))\). So, \(\mathbf{v}(\frac{1}{\sqrt{t}}) = \left\langle (\frac{1}{\sqrt{t}})^2, -2(\frac{1}{\sqrt{t}}), 1 \right\rangle = \left\langle \frac{1}{t}, -\frac{2}{\sqrt{t}}, 1 \right\rangle\) Now, we apply the chain rule: \(\frac{d\mathbf{v}(g(t))}{dt} = \left\langle \frac{d(\frac{1}{t})}{dt}, \frac{d(-\frac{2}{\sqrt{t}})}{dt}, \frac{d(1)}{dt} \right\rangle \cdot \frac{dg(t)}{dt}\) Differentiating each of the component functions of the vector, we get: \(\frac{d\mathbf{v}(g(t))}{dt} = \left\langle -\frac{1}{t^2}, \frac{1}{t\sqrt{t}}, 0 \right\rangle \cdot \frac{1}{\sqrt{t}}\) Now, multiplying each component of the resulting vector by \(\frac{dg(t)}{dt}\), we find the derivative of \(\mathbf{v}(g(t))\): \(\frac{d\mathbf{v}(g(t))}{dt} = \left\langle -\frac{1}{t^{5/2}}, \frac{1}{t^{3/2}}, 0 \right\rangle\) The derivative of the function \(\mathbf{v}(g(t))\) is: $$\frac{d\mathbf{v}(g(t))}{dt} = \left\langle -\frac{1}{t^{5/2}}, \frac{1}{t^{3/2}}, 0 \right\rangle$$