Find the equations of the tangent and the normal to the curve y = 1 + cos x at the point where x = pi/6. Show that the area enclosed by the tangent, normal and the x-axis is 5/4 (1 + sq.3 /2)^2
No problem getting the equations. However, I do not understand the workings and concept behind this area thing. How can I make use the the equations I have to derive the area?
Thank you.
this must be Qn 8 from Panpac A Maths
U had eqns of normal ,tangent and x-axis
the three axis form a triangle
the normal and tangent intercept at a point, that is the upper vertice of the triangle
the normal and tangent intercept x-axis is the coordinates of the base length of the triangle
find length of height and base
den using triangle formulae, find area
Thanks SBS n SMRT (:
The volume of water in a container V (cm3) after t seconds is given by V = 20e^-1/5. Find, in terms of e, the rate of change of V when t = 5.
Answer is -4/e but I got 4/e. Would need the full solution for this question if possible... Anyway, this question doesn't require me to rearrange a formula of my own, does it? (eg. dy/dt = dy/dx X dx/dt)
Hi,
-4/e is correct. Do check your working. There is no need to use chain rule. Thanks!
Cheers,
Wen Shih
Hi. I still can't get the answers for both questions somehow.... is it possible to show me the workings? thanks. i really need help for the first question.
*Edited: Managed to find the answer to question 2. (: I only need help in solving question 1.
Hi,
Draw the triangle so that you can see what the height and base of the triangle are.
Thanks!
Cheers,
Wen Shih
Originally posted by anpanman:Thanks SBS n SMRT (:
The volume of water in a container V (cm3) after t seconds is given by V = 20e^-1/5. Find, in terms of e, the rate of change of V when t = 5.
Answer is -4/e but I got 4/e. Would need the full solution for this question if possible... Anyway, this question doesn't require me to rearrange a formula of my own, does it? (eg. dy/dt = dy/dx X dx/dt)
welcme.....