Question 27.4: A curve, C, is defined parametrically by x = 4 y =t³ z = 5+t......
A curve, C, is defined parametrically by
x = 4 y =t³ z = 5+t
and is located within a vector field F = yi + x²j + (z + x)k.
(a) Find the coordinates of the point P on the curve where the parameter t takes the value 1.
(b) Find the coordinates of the point Q where the parameter t takes the value 3.
(c) By expressing the line integral \int_{C}\mathbf{F}\cdot\mathrm{d}\mathbf{s} entirely in terms of t find the value of the line integral from P to Q. Note that ds = dxi + dyj + dzk.
(a) When t = 1, x = 4, y = 1 and z = 6, and so P has coordinates (4, 1, 6).
(b) When t = 3, x = 4, y = 27 and z = 8, and so Q has coordinates (4, 27, 8).
(c) To express the line integral entirely in terms of t we note that if x = 4, {\frac{\mathrm{d}x}{\mathrm{d}t}}= 0 so that dx is also zero. If y = t³ then {\frac{\mathrm{d}y}{\mathrm{d}t}}\;=\;3t^{2} so that dy = 3t² dt.
Similarly since z=5+t,{\frac{\mathrm{d}z}{\mathrm{d}t}}=1 so that dz = dt. The line integral becomes
\int_{C}\mathbf{F}\cdot\mathrm{d}\mathbf{s}=\int_{C}(y\mathbf{i}+x^{2}\mathbf{j}+(z+x)\mathbf{k})\cdot(\mathrm{d}x\mathbf{i}+\mathrm{d}y\mathbf{i}+\mathrm{d}z\mathbf{k})\\ = {{{\int}}}_{C}{y} d x+x^{2}dy+(z+x)dz\\ =\int_{t=1}^{t=3}0+16(3t^{2})\,\mathrm{d}t+(9+t)\,\mathrm{d}t\\ =\int_{1}^{3}48t^{2}+9+t\,\mathrm{d}t\\ =\left[{\frac{48t^{3}}{3}}+9t+{\frac{t^{2}}{2}}\right]^3_1\\ =\left({\frac{48(3^{3})}{3}}+(9)(3)+{\frac{3^{2}}{2}}\right)-\left({\frac{48}{3}}+9+{\frac{1}{2}}\right)\\= 438