e993d6…: “Thanks to a heavy solution”

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Consider the graph below of the vector-valued function $\vec {p}(s)$ . Additionally, suppose that $\vec {p}$ is parameterized by arc length.

Compute: $\int _0^{10} \vec {p}'(s)\dotp \vec {p}'(s)\d s$ $\int _0^{10} \vec {p}'(s)\dotp \vec {p}'(s)\d s =\answer {10}$

Since the curve uses arclength as a parameter, we must have that $\left |\vec {p}'(s)\right | = \answer {1}$ . How is the quantity $\vec {p}'(s)\dotp \vec {p}'(s)$ related to $\left |\vec {p}'(s)\right |$ ?

Compute: $\vec {p}(3)$

$\vec {p}(3) = \vector {\answer {3},\answer {-1}}$

Since we are given that the curve uses arclength as a parameter, $\vec {p}(3)$ can be found from the $x$ and $y$ coordinates after we travel $3$ units along the curve. After traveling $2$ units, we arrive at $(3,-2)$ , and traveling $1$ more unit brings us to $(3,-1)$ .

Compute: $\vec {p}(7.5)$

$\vec {p}(7.5) = \vector {\answer {1},\answer {-.5}}$

Compute $\eval {\dd {s} \left (\vec {p}(s)\dotp \vec {p}(s)\right )}_{s=0}$

$\eval {\dd {s} \left (\vec {p}(s)\dotp \vec {p}(s)\right )}_{s=0} = \answer {-10}$

Using the formula for differentiating a dot product,

$\dd {s} \left (\vec {p}(s)\dotp \vec {p}(s)\right ) = \vec {p}'(s)\dotp \vec {p}(s) + \vec {p}(s)\dotp \vec {p}'(s) = 2 \left (\vec {p}'(s)\dotp \vec {p}(s)\right ).$

For $0 \leq s \leq 2$ , note that the curve is a horizontal line, and a tangent vector with unit speed in the direction of the curve is $\vector {\answer {-1},\answer {0}}$ , so $\vec {p}'(0) = \vector {\answer {-1},\answer {0}}$ .

Also, $\vec {p}(0) = \vector {\answer {5},\answer {-2}}$ , so

$\eval {\dd {s} \left (\vec {p}(s)\dotp \vec {p}(s)\right )}_{s=0} = 2 \left (\vec {p}'(0)\dotp \vec {p}(0)\right ).$