Equation of tangent line to the curve

[Win_odd Dhamnekar]

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.

 

[Dr.Peterson]

But answer provided to me is different. How is that? I want to know where i am wrong?

I can't say why the provided answer is different without seeing what it is. Maybe it is equivalent to yours.

 

[Win_odd Dhamnekar]

I can't say why the provided answer is different without seeing what it is. Maybe it is equivalent to yours.

Hi,
The answer provided to me is $x-\frac{a}{\sqrt{2}}=y-\frac{a}{\sqrt{2}}=\frac{\left(z-a\frac{\pi}{4}\tan{\alpha}\right)}{\sqrt{2}\tan{\alpha}}$ at $\theta=\frac{\pi}{4}$

 

[Dr.Peterson]

That answer is equivalent to yours apart from a sign. (Multiply all sides of yours by [MATH]\frac{a}{\sqrt{2}}[/MATH].)

Which of the two do you think is right?

 

[Win_odd Dhamnekar]

That answer is equivalent to yours apart from a sign. (Multiply all sides of yours by [MATH]\frac{a}{\sqrt{2}}[/MATH].)

Which of the two do you think is right?

Hi,
Answer provided to me have positive x-coordinate in the equation of tangent line because $\sin{\theta}$ can not be negative for $\theta=\frac{\pi}{4}$. I didn't notice it, that'why i put negative sign for x-coordinate of tangent line equation.

I think this justification is sufficient.

 

[Dr.Peterson]

I say your answer is right and theirs, as you quoted it, is wrong. If you graph the curve (a helix), you can see this.

 

[Win_odd Dhamnekar]

I say your answer is right and theirs, as you quoted it, is wrong. If you graph the curve (a helix), you can see this.

Hi,
How can i plot this question and its solution in my graphing calculator 3D? I tried to plot the curve and its tangent line at $\theta=\frac{\pi}{4}$ using following graphic controls. $a=1, 0\leq \theta \leq 2*\pi$ or $\frac{\pi}{2}, 0\leq \alpha\leq 2*\pi$ or $\frac{\pi}{2}$. But i could not identify which is the curve and which is the tangent line.
Would you tell me how to plot, view and interpret this curve and its tangent line graphically?

 

[Dr.Peterson]

I just plotted it in my head! I don't have a graphing calculator to try it on. There are online tools you could try, that can do it in color.