-2
$\begingroup$

I have tried this question more than a dozen times now. The question is find the point(s) at which the line normal to $y = 2\arcsin 0.5x$ is parallel to the line $y = 1-x$.

Edit: I changed y= 2arcsin0.5x to y=1/2arcsin0.5x then derive it. Which is wrong. Thanks for the answers.

$\endgroup$
  • $\begingroup$ have you tried doing it? $\endgroup$ – Pushkar Soni Jan 22 '17 at 4:47
  • $\begingroup$ Yes I did and i wasn't doing the right steps. Check edit. $\endgroup$ – n.j Jan 22 '17 at 5:11
  • $\begingroup$ just do the same procedure, you'll get the required equation. $\endgroup$ – Pushkar Soni Jan 22 '17 at 5:15
  • 1
    $\begingroup$ however, the site is not meant for the purpose of homework questions. when you ask a question please provide the solution you have done. if you are having a problem with a particular step. people will help, otherwise not. and please do not expect complete answers to these questions this site will only provide you a hint. so you could solve it yourself. $\endgroup$ – Pushkar Soni Jan 22 '17 at 5:20
  • $\begingroup$ i guess you are having problem regarding differentiation.right? $\endgroup$ – Pushkar Soni Jan 22 '17 at 5:22
1
$\begingroup$

we have, $$ y= arcsin(0.5)$$ differentiate it to get

$$ \frac{1}{\sqrt{(1-0.25x^2)}} $$ which is ofcourse the slope of the tangent.

then the slope of the normal is given by:

$$ -\frac{\sqrt{(1-0.25x^2)}}{1} $$

for second equation

$$y=1-x$$

the slope is $$-1$$

according to your question both the slope of normal and slope of the line should be equal since they are parallel.

equate both the equation to get the required result.

$\endgroup$
0
$\begingroup$

Hint:

$$y=2\arcsin 0.5 x$$

$$\frac{dy}{dx}=\frac{2(0.5)}{\sqrt{1-(0.5x)^2}} $$

The gradient of the normal line would be $-\sqrt{1-(0.5x)^2}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.