1. ## [SOLVED] Advanced parametric diff question.

A curve is defined by the parametric equations x = acos^2 t, y = asin^3 t
for range 0<= t <= pi/2 where a is a positive constant.

(i) Find and simplify an expresion for dy/dx in terms of t

(ii) Find the equation of the noraml at the point on the curve where
t = pi/6

(ii) If the normal cuts the the axis at point M and N, find the area of the triangle of the triangle OMN, in terms of a, where O is the origin.

OK.

(i)after finding dx/dt and dy/dt I simplfy to get -3/2(sin t)

(ii) The grad. of tangent is -3/4 when t = pi/6. Therefore the normal will be 4/3

(iii) I'm lost with this part...

Am I correct up to (iii) and can anyone help???

Thanks

2. Originally Posted by dojo
A curve is defined by the parametric equations x = acos^2 t, y = asin^3 t
for range 0<= t <= pi/2 where a is a positive constant.

(i) Find and simplify an expresion for dy/dx in terms of t

(ii) Find the equation of the noraml at the point on the curve where
t = pi/6

(ii) If the normal cuts the the axis at point M and N, find the area of the triangle of the triangle OMN, in terms of a, where O is the origin.

OK.

(i)after finding dx/dt and dy/dt I simplfy to get -3/2(sin t)

(ii) The grad. of tangent is -3/4 when t = pi/6. Therefore the normal will be 4/3

(iii) I'm lost with this part...

Am I correct up to (iii) and can anyone help???

Thanks
find the equation of the normal line ... you have its slope, all you need is the point on the curve at $t = \frac{\pi}{6}$

once you get the normal line equation, determine its intercepts ... the rest should be downhill from there.

3. Originally Posted by dojo
A curve is defined by the parametric equations x = acos^2 t, y = asin^3 t
for range 0<= t <= pi/2 where a is a positive constant.

(i) Find and simplify an expression for dy/dx in terms of t

(ii) Find the equation of the normal at the point on the curve where
t = pi/6

(iii) If the normal cuts the the axis at point M and N, find the area of the triangle of the triangle OMN, in terms of a, where O is the origin.

OK.

(i)after finding dx/dt and dy/dt I simplfy to get -3/2(sin t)

(ii) The grad. of tangent is -3/4 when t = pi/6. Correct up to here.
Therefore the normal will be 4/3 The slope of the normal will be 4/3. But the question asks for the equation of the normal.
The equation of the normal will be $y-y_0 = \tfrac43(x-x_0)$, where $(x_0,y_0)$ is the point on the curve given by $t = \pi/6$, in other words the point $\bigl(a\cos^2\tfrac\pi6,a\sin^3\tfrac\pi6\bigr)$.

Once you know the equation for the normal, you should be able to find the points M and N where it crosses the axes. That will tell you the lengths OM and ON, and then you can calculate the area of the triangle OMN as $\tfrac12OM\times ON$.

4. Thank you both for your comments. I've attempted this again and wondered if you could check it for me. I'm also confused about the "a" the positive constant, as I dont know this I am right to expect it in my answer?

Here is what I've done:

x = acos^2 t, y = asin^3 t
... so when t = pi/6
x = 3a/4
y = a/8

Putting these in the equation of the line:

y - (a/8) = 4/3(x - (3a/4))
24y = 32x - 21a
With some manipulation when x, y are set to zero the intercepts can be found:
x = 0 then y = -7a/8 (call this point M)
y = 0 then x = 21a/32 (call this point N)

So the lenghts OM = 7a/8 and ON = 21a/32

Therefore the area is 1/2*(7a/8 * 21a/32) which is (147a^2)/512.

Is this right?

Thanks, D

x = 0 the y intercept

5. Originally Posted by dojo
Thank you both for your comments. I've attempted this again and wondered if you could check it for me. I'm also confused about the "a" the positive constant, as I dont know this I am right to expect it in my answer?

Here is what I've done:

x = acos^2 t, y = asin^3 t
... so when t = pi/6
x = 3a/4
y = a/8

Putting these in the equation of the line:

y - (a/8) = 4/3(x - (3a/4))
24y = 32x - 21a
With some manipulation when x, y are set to zero the intercepts can be found:
x = 0 then y = -7a/8 (call this point M)
y = 0 then x = 21a/32 (call this point N)

So the lenghts OM = 7a/8 and ON = 21a/32

Therefore the area is 1/2*(7a/8 * 21a/32) which is (147a^2)/512.

Is this right?

Thanks, D

x = 0 the y intercept
looks fine to me.