Calculus problem #1994

$$ Along the curve $C$ , we have parametric equations: $$ $\begin{aligned} \langle x,y,z\rangle &=\langle 2,0,0\rangle + t\langle 1,4,5\rangle\;\;\; \text{for} \;\; 0\le t\le 1.\\ x&=2+t \;\;\;\Rightarrow\;\;\; dx=dt,\\ y&=4t \;\;\;\;\;\;\;\Rightarrow\;\;\; dy=4\,dt,\\ z&=5t \;\;\;\;\;\;\;\Rightarrow\;\;\; dz=5\,dt.\\ \end{aligned}$ Therefore $\begin{aligned} N &=\int_0^1 \left(4t\,dt+5t\,(4\,dt)+(2+t)\,(5\,dt)\right)\\ &=\int_0^1 (29t+10)\,dt\\ &=\left.\left(\frac{29}{2}t^2+10t\right)\right|_0^1\\ &=\boxed{24.5} \end{aligned}$ $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

We'll change the integral from a multi-variable to a single variable one. This can be easily done by establishing a relation between the variables $x,y,z$ . We see the integration is over the line-segment $C$ which joins $(2,0,0)$ and $(3,4,5)$ . So $x\in(2,3),y\in(0,4),z\in(0,5)$ . Extending the concept of straight line in 2 $-D$ to 3 $-D$ , we have

$\displaystyle\frac{x-2}{3-2}=\frac{y-0}{4-0}=\frac{z-0}{5-0}$ .

Now the task which we set out to do. Using the above relationship we have the integral as

$\int\limits_2^3 4(x-2)\mathrm{d}x+5(x-2)\cdot4\mathrm{d}x+x\cdot5\mathrm{d}x=\frac{-27}{2}+38=\boxed{24.5}$