Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

tangents and normals find the equation of the tangent to the curve x= sin 3t , y=cos 2t at t=pi/4

tangents and normals
find the equation of the tangent to the curve x= sin 3t , y=cos 2t at t=pi/4

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:
Hello Student,
Please find answer to your question,

x = sin3t
dx = 3cos3t.dt…..........(1)
y = cos2t
dy = -2sin2t.dt…..........(2)
Divide (2) by (1)
\frac{dy}{dx} = \frac{-2sin2t}{3cos3t}
t = \frac{\pi }{4}
\frac{dy}{dx} = \frac{-2sin2\frac{\pi }{4}}{3cos3\frac{\pi }{4}}
\frac{dy}{dx} = \frac{2\sqrt{2}}{3}
This is slope of tangent.
Point is
(\frac{1}{\sqrt{2}}, 0)
Equation is:
(y-0) = \frac{2\sqrt{2}}{3}(x - \frac{1}{\sqrt{2}})
y = \frac{2\sqrt{2}}{3}(x - \frac{1}{\sqrt{2}})


Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free