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You can plot this using GC in exams, but for visualisation, desmos.com is very useful. Try it on your own someday! I will explain how to get the intercepts and part (ii) in the next post.
Date Posted:
3 years ago
But when i plotted this into my gc, i only got the bottom part of the curve from x=0 onwards, so i was very confused haha
Maybe you can take a photo of it and post it here?
Sometimes it might be the settings of the GC, eg. Scale and zoom in/out
Sometimes it might be the settings of the GC, eg. Scale and zoom in/out
Okies, i just added the pic of the gc as an answer under this qns! And also sorry for the late reply
On the y-axis, x = 0
e^(sin(t)) - 1 = 0
e^(sin(t)) = 1
sin(t) = ln 1 = 0
t = 0, π, 2π,...
So y = e^(-cos 0) - 1, e^(-cos π) - 1, e^(-cos 2π) - 1
y = e-¹ - 1 , e¹ - 1 , e-¹ - 1 ,...
y = 1/e - 1 , e - 1 , 1/e - 1,...
For x = 0, the values of y alternate like how your cos x and sin x curves oscillate. There are only 2 unique values of y.
So the coordinates of the y-intercepts are (0, e - 1) and (0, 1/e - 1)
Likewise,
On the x-axis, y = 0
e^(-cos(t)) - 1 = 0
e^(-cos(t)) = 1
-cos(t) = ln 1 = 0
cos(t) = 0
t = cos-¹(0) = π/2, 3π/2, 5π/2,...
So x = e^(sin π/2) - 1, e^(sin 3π/2) - 1, e^(sin 5π/2) - 1
x = e¹ - 1, e-¹ - 1, e¹ - 1, ...
x = e - 1 , 1/e - 1 , e - 1,...
For y = 0, the values of x alternate like how your cos x and sin x curves oscillate. There are only 2 unique values of x.
So the coordinates of the x-intercepts are (e - 1, 0) and (1/e - 1, 0)
e^(sin(t)) - 1 = 0
e^(sin(t)) = 1
sin(t) = ln 1 = 0
t = 0, π, 2π,...
So y = e^(-cos 0) - 1, e^(-cos π) - 1, e^(-cos 2π) - 1
y = e-¹ - 1 , e¹ - 1 , e-¹ - 1 ,...
y = 1/e - 1 , e - 1 , 1/e - 1,...
For x = 0, the values of y alternate like how your cos x and sin x curves oscillate. There are only 2 unique values of y.
So the coordinates of the y-intercepts are (0, e - 1) and (0, 1/e - 1)
Likewise,
On the x-axis, y = 0
e^(-cos(t)) - 1 = 0
e^(-cos(t)) = 1
-cos(t) = ln 1 = 0
cos(t) = 0
t = cos-¹(0) = π/2, 3π/2, 5π/2,...
So x = e^(sin π/2) - 1, e^(sin 3π/2) - 1, e^(sin 5π/2) - 1
x = e¹ - 1, e-¹ - 1, e¹ - 1, ...
x = e - 1 , 1/e - 1 , e - 1,...
For y = 0, the values of x alternate like how your cos x and sin x curves oscillate. There are only 2 unique values of x.
So the coordinates of the x-intercepts are (e - 1, 0) and (1/e - 1, 0)
I will post part (ii) here since the app keeps hanging whenever I try to post it as a separate answer.
(ii)
y = e^(-cos t) - 1
dy/dt = (sin t) e^(-cos t) = e^(-cos t) sin t
x = e^(sin t) - 1
dx/dt = (cos t)e^(sin t) = e^(sint) cos t
Now dy/dx × dx/dt = dy/dt
So dy/dx × e^(sint) cos t = e^(-cos t) sin t
dy/dx = (e^(-cos t) sin t) / (e^(sint) cos t)
= e^(-cos t - sin t) (sin t / cos t)
= e^(-cos t - sin t) tan t
When t = 3π/4,
dy/dx
= e^(-cos 3π/4 - sin 3π/4) tan (3π/4)
= e^(-(-√2 / 2) - √2 / 2)) (-1)
= -e^(√2 / 2 - √2 / 2)
= -e^0
= -1
So gradient of tangent = -1
Gradient of normal = -1/-1 = 1
When t = 3π/4,
y = e^(-cos 3π/4) - 1
= e^(-(-√2 / 2)) - 1
= e^(√2 / 2) - 1
x = e^(sin 3π/4) - 1
= e^(√2 / 2) - 1
Notice that x = y.
Sub x,y = e^(√2 / 2) - 1 and m = 1 into y = mx + c, (equation of line)
e^(√2 / 2) - 1 = 1(e^(√2 / 2) - 1) + c
c = 0
Therefore, equation of the normal to curve C at point P is y = x (shown)
Lastly, since we have established that the equation of the normal is y = x ,
then at the other intersection point of the normal and the curve C, y = x as well.
So e^(-cos t) - 1= e^(sin t) - 1
e^(-cos t) = e^(sin t)
-cos t = sin t
sin t / cos t = -1
tan t = -1
t = tan-¹ (-1) = -π/4, π - π/4, 2π - π/4 ,...
t = -π/4, 3π/4, 7π/4,...
Since 0 ≤ t ≤ 2π and we know t = 3π/4 was for point P,
then t = 7π/4 at the other intersection point.
x = e^(-cos 7π/4) - 1 = e^(-√2 / 2) - 1
y = x = e^(-√2 / 2) - 1
∴The coordinates of the other intersection point are ( e^(-√2 / 2) - 1, e^(-√2 / 2) - 1 )
What we notice about this curve is that : it is symmetrical about the line y = x. (Realise that each y-intercept maps exactly onto one x-intercept, and vice versa)
(ii)
y = e^(-cos t) - 1
dy/dt = (sin t) e^(-cos t) = e^(-cos t) sin t
x = e^(sin t) - 1
dx/dt = (cos t)e^(sin t) = e^(sint) cos t
Now dy/dx × dx/dt = dy/dt
So dy/dx × e^(sint) cos t = e^(-cos t) sin t
dy/dx = (e^(-cos t) sin t) / (e^(sint) cos t)
= e^(-cos t - sin t) (sin t / cos t)
= e^(-cos t - sin t) tan t
When t = 3π/4,
dy/dx
= e^(-cos 3π/4 - sin 3π/4) tan (3π/4)
= e^(-(-√2 / 2) - √2 / 2)) (-1)
= -e^(√2 / 2 - √2 / 2)
= -e^0
= -1
So gradient of tangent = -1
Gradient of normal = -1/-1 = 1
When t = 3π/4,
y = e^(-cos 3π/4) - 1
= e^(-(-√2 / 2)) - 1
= e^(√2 / 2) - 1
x = e^(sin 3π/4) - 1
= e^(√2 / 2) - 1
Notice that x = y.
Sub x,y = e^(√2 / 2) - 1 and m = 1 into y = mx + c, (equation of line)
e^(√2 / 2) - 1 = 1(e^(√2 / 2) - 1) + c
c = 0
Therefore, equation of the normal to curve C at point P is y = x (shown)
Lastly, since we have established that the equation of the normal is y = x ,
then at the other intersection point of the normal and the curve C, y = x as well.
So e^(-cos t) - 1= e^(sin t) - 1
e^(-cos t) = e^(sin t)
-cos t = sin t
sin t / cos t = -1
tan t = -1
t = tan-¹ (-1) = -π/4, π - π/4, 2π - π/4 ,...
t = -π/4, 3π/4, 7π/4,...
Since 0 ≤ t ≤ 2π and we know t = 3π/4 was for point P,
then t = 7π/4 at the other intersection point.
x = e^(-cos 7π/4) - 1 = e^(-√2 / 2) - 1
y = x = e^(-√2 / 2) - 1
∴The coordinates of the other intersection point are ( e^(-√2 / 2) - 1, e^(-√2 / 2) - 1 )
What we notice about this curve is that : it is symmetrical about the line y = x. (Realise that each y-intercept maps exactly onto one x-intercept, and vice versa)
Oh i see! Thank u so much for the explanation it was very clear :)
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Heres the pic of the gc!
Date Posted:
3 years ago
Hmm strange. I've tried it using TI-84 Plus and it works fine.
Maybe you'll have to press the ZoomStandard / Zoom Square function
Maybe you'll have to press the ZoomStandard / Zoom Square function
Ahh i see! Thank u!