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Parametric form of a curve, cartesian points, tangent vector, and speed.

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You are given the following curve, $t\\mapsto\\pmatrix{t^\\var{a},\\simplify{{b}t}}$, defined with respect to the parameter $t$.

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a)

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To find the coordinates of the point corresponding to $t=\\var{c}$, substitute $t=\\var{c}$ into the expression for the curve, i.e.

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\\[\\pmatrix{x,y}=\\pmatrix{\\var{c}^\\var{a},\\var{b}\\times\\var{c}}=\\pmatrix{\\var{x},\\var{y}}.\\]

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b)

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Differentiate each component of the vector in part a) to find the tangent vector $\\boldsymbol{u}$, i.e.

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\\[\\boldsymbol{u}=\\pmatrix{\\frac{\\mathrm{d}}{\\mathrm{d}t}t^\\var{a},\\frac{\\mathrm{d}}{\\mathrm{d}t}\\simplify{{b}t}}=\\pmatrix{\\var{a}t^\\var{a-1},\\var{b}}.\\]

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The tangent vector at $t=\\var{c}$ is found by substituting $t=\\var{c}$ into the tangent vector $\\boldsymbol{u}$, i.e.

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\\[\\boldsymbol{u}\\vert_{t=\\var{c}}=\\pmatrix{\\var{a}\\times\\var{c}^\\var{a-1},\\var{b}}=\\pmatrix{\\var{dxdtc},\\var{b}}.\\]

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c)

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The velocity $u$ is given by $u=\\lvert\\boldsymbol{u}\\rvert=\\sqrt{\\left(\\frac{\\mathrm{d}x}{\\mathrm{d}t}\\right)^2+\\left(\\frac{\\mathrm{d}y}{\\mathrm{d}t}\\right)^2}$.  We must calculate the speed at $t=\\var{d}$, however, therefore

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\\[u\\vert_{t=\\var{d}}=\\sqrt{\\left(\\var{a}\\times\\var{d}^\\var{a-1}\\right)^2+\\var{b}^2}=\\sqrt{\\var{dxdtd^2}+\\var{b^2}}=\\var{speed} \\; \\text{to 2d.p.}\\]

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Find the coordinates $\\pmatrix{x,y}$ of the point corresponding to $t=\\var{c}$.

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$\\pmatrix{x,y}=($[[0]]$,$[[1]]$)$.

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Find the components of the tangent vector $\\boldsymbol{u}\\equiv\\pmatrix{\\frac{\\mathrm{d}x}{\\mathrm{d}t},\\frac{\\mathrm{d}y}{\\mathrm{d}t}}$.

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$\\boldsymbol{u}=($[[0]]$,$[[1]]$)$.

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The components of the same tangent vector, given $t=\\var{c}$.

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$\\boldsymbol{u}|_{t=\\var{c}}=($[[2]]$,$[[3]]$)$.

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Interpreting $t$ as time, and hence the tangent vector $\\boldsymbol{u}$ as velocity, find the speed $u=|\\boldsymbol{u}|$ at $t=\\var{d}$.

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$u=$ [[0]].  (Enter your answer to 2d.p.)

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