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Chapter 11: Problem 15

Find a tangent vector at the given value of \(t\) for the following parameterized curves. $$\mathbf{r}(t)=\left\langle t, 3 t^{2}, t^{3}\right\rangle, t=1$$

Short Answer

Expert verified
Question: Find the tangent vector to the curve \(\mathbf{r}(t)=\left\langle t, 3 t^{2}, t^{3}\right\rangle\) at t = 1. Answer: The tangent vector at t = 1 is \(\left\langle 1, 6, 3\right\rangle\).

Step by step solution

01

Find the Derivative of the Parameterized Curve

The given parameterized curve is: $$\mathbf{r}(t)=\left\langle t, 3 t^{2}, t^{3}\right\rangle$$ To find the derivative of this curve, we will find the derivative of each component with respect to t. $$\frac{d\mathbf{r}}{dt}=\left\langle \frac{dt}{dt}, \frac{d(3t^2)}{dt}, \frac{d(t^3)}{dt}\right\rangle$$
02

Evaluate the Derivative

Now, let's compute the derivatives of each component. $$\frac{dt}{dt} = 1$$ $$\frac{d(3t^2)}{dt} = 6t$$ $$\frac{d(t^3)}{dt} = 3t^2$$ So, the derivative of the parameterized curve is: $$\frac{d\mathbf{r}}{dt} = \left\langle 1, 6t, 3t^2 \right\rangle$$
03

Evaluate the Derivative at t = 1

Now, we are going to find the tangent vector at t = 1. Hence, substitute t = 1 into the derivative. $$\frac{d\mathbf{r}}{dt}|_{t=1} = \left\langle 1, 6(1), 3(1^2) \right\rangle = \left\langle 1, 6, 3\right\rangle$$
04

State the Answer

The tangent vector to the curve \(\mathbf{r}(t)=\left\langle t, 3 t^{2}, t^{3}\right\rangle\) at t = 1 is \(\left\langle 1, 6, 3\right\rangle\).

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