Win_odd Dhamnekar
Junior Member
- Joined
- Aug 14, 2018
- Messages
- 207
Hi,
How to find the equation of tangent line to the curve \(x=a\cos{\theta}, y=a\sin{\theta},z=a\theta \tan{\alpha}\) at \(\frac{\pi}{4}\)?
Solution:- Given curve is \(\vec{r}= a\cos{\theta}\hat{i}+a\sin{\theta}\hat{j}+a\theta\tan{\alpha}\hat{k}\),
Differentiating w.r.t \(\theta, \frac{d\vec{r}}{d\theta}=-a\sin{\theta}\hat{i}+a\cos{\theta}\hat{j}+a\tan{\alpha}\hat{k}\)
So the equation of a tangent line to the curve \(\vec{r}\) is \(\frac{x-\frac{a}{\sqrt{2}}}{-\frac{a}{\sqrt{2}}}= \frac{y-\frac{a}{\sqrt{2}}}{\frac{a}{\sqrt{2}}}=\frac{z-a\frac{\pi}{4}\tan{\alpha}}{a\tan{\alpha}}\) at \(\theta =\frac{\pi}{4}\)
But answer provided to me is different. How is that? I want to know where i am wrong?
If any member knows the answer to this question may reply with correct answer.
How to find the equation of tangent line to the curve \(x=a\cos{\theta}, y=a\sin{\theta},z=a\theta \tan{\alpha}\) at \(\frac{\pi}{4}\)?
Solution:- Given curve is \(\vec{r}= a\cos{\theta}\hat{i}+a\sin{\theta}\hat{j}+a\theta\tan{\alpha}\hat{k}\),
Differentiating w.r.t \(\theta, \frac{d\vec{r}}{d\theta}=-a\sin{\theta}\hat{i}+a\cos{\theta}\hat{j}+a\tan{\alpha}\hat{k}\)
So the equation of a tangent line to the curve \(\vec{r}\) is \(\frac{x-\frac{a}{\sqrt{2}}}{-\frac{a}{\sqrt{2}}}= \frac{y-\frac{a}{\sqrt{2}}}{\frac{a}{\sqrt{2}}}=\frac{z-a\frac{\pi}{4}\tan{\alpha}}{a\tan{\alpha}}\) at \(\theta =\frac{\pi}{4}\)
But answer provided to me is different. How is that? I want to know where i am wrong?
If any member knows the answer to this question may reply with correct answer.