Answer:
(h) y = -8/3·x + 16·a/3
(i) dt = √(x² + y²) = √( (a + 3·t)² + (b + 8·t)²)
(j) [tex]\dfrac{d(t)}{dt} = \dfrac{3\cdot a +8\cdot b+ 73 \cdot t}{\sqrt{a^2 + 6\cdot a \cdot t + b^2+16 \cdot b \cdot t + 73\cdot t} }[/tex]
Step-by-step explanation:
(h) The location of the object = P(t) = (x(t), y(t))
Where;
x(t) = a + 3·t
y(t) = b + 8·t
Therefore we have;
At x = 0, t = -a/3. at y = 0, b = -8·t = -8·a/3
y = -8·a/3 + 8·t = -8/3(a - 3·t) = -8/3(x - 2·a) = -8/3·x + 16·a/3
y = -8/3·x + 16·a/3
(i) dt = √(x² + y²) = √( (a + 3·t)² + (b + 8·t)²)
(j) The instantaneous rate of change of d(t), with a corresponding change in time, t, obtained by solving through an online platform, is given as follows;
[tex]\dfrac{d(t)}{dt} = \dfrac{d}{dt} \left (\sqrt{\left (a + 3\cdot t \right)^2 + \left (b + 8\cdot t \right)^2} \right ) = \dfrac{3\cdot a +8\cdot b+ 73 \cdot t}{\sqrt{a^2 + 6\cdot a \cdot t + b^2+16 \cdot b \cdot t + 73\cdot t} }[/tex].
