DE and vector question (1 Viewer)

kractus · Sunday at 7:21 PM

Screen Shot 2023-03-19 at 7.21.00 pm.png

Screen Shot 2023-03-19 at 7.21.24 pm.png

How would you do these?

jimmysmith560 · Sunday at 7:49 PM

Would the following working help for the first question?

$\Rightarrow \frac{R_{A}}{R_{B}}= \frac{\left ( T_{tA}-T_{s} \right )}{\left ( T_{tB}-T_{s} \right )}$

$\Rightarrow \frac{R_{A}}{R_{B}}= \frac{\left ( T_{tA}-T_{s} \right )}{\left ( T_{tB}-T_{s} \right )}$

$\Rightarrow \frac{R_{A}}{R_{B}}= \frac{\left ( 120-40 \right )}{\left ( 80-40 \right )}$

$\Rightarrow \frac{R_{A}}{R_{B}}= \frac{2}{1} \ \text{i.e. (B).}$

kractus · Tuesday at 6:09 PM

Thanks it helped. What about the other question? I'm still kinda confused lol

jimmysmith560 · Tuesday at 7:01 PM

kractus said:
Thanks it helped. What about the other question? I'm still kinda confused lol

No worries. Would the following help with the other question?

$x'(t)=\frac{-1}{\sqrt{1-t^2}}$

$y'\left(t\right)=\frac{1}{\sqrt{1-t^2}}$

$\text{Distance}=\int _{-1}^1\sqrt{\left(\frac{-1}{\sqrt{1-t^2}}\right)^2+\left(\frac{-1}{\sqrt{1-t^2}}\right)^2}dt$

$=\sqrt{2}\int _{-1}^1\frac{1}{\sqrt{1-t^2}}\ dt=\sqrt{2}\pi \: \ \text{i.e. (D).}$

DE and vector question (1 Viewer)

kractus

Member

jimmysmith560

Le Phénix Trilingue

kractus

Member

jimmysmith560

Le Phénix Trilingue

Users Who Are Viewing This Thread (Users: 0, Guests: 1)