Need Calculus help ASAP!!?

Question

I'm a junior in a college calculus class. The following problem
was assigned, but the next day, after no one had completed it,
our professor attempted to do it but quit after 25 minutes.

Find the equation of the line tangent to the ellipse
b^2*x^2 + a^2*y^2 = a^2*b^2
in the first quadrant that forms with the coordinate axes
the triangle of smallest possible area
(a & b are positive constants)

Best Answer Gets 10 points!

Answer 1

Best answer gets 10 points, and you no doubt get a prize from the lecturer? Is that fair? :)

Anyway.
Differentiate implicitly to find the gradient:
2xb^2 + 2yy'a^2 = 0
y' = (-xb^2)/(ya^2).

Thus, the equation of the tangent at any point (a cos t, b sin t) is:
y - b sin t = y'(x - a cos t).

Now, we find the x and y intercepts:
When x = 0, we get y = b sin t- ay' cos t.
When y = 0, we get x = (-b sin t)/y' + a cos t.

Thus, the area of the triangle is 0.5(b sin t- ay' cos t)((-b sin t)/y' + a cos t).

Now, take a second to shudder, and lets dive into differentiating it.
I'm going to ignore the 0.5 on the front, it doesn't affect anything.
Expand it:
-b^2 (sin t)^2 / y' +ab sin t cos t + ab sin t cos t - a^2 y' (cos t)^2

Simplify:
2ab sin t cos t - b^2(sin t)^2 / y' - a^2 (cos t)^2 * y'.
Must be about time to substitute in for y'.
2ab sin t cos t - b^2(sin t)^2 * (b sin t * a^2)/(-a cos t * b^2) - a^2 (cos t)^2 * (-a cos t * b^2)/(b sin t * a^2).

Simplify again:
2ab sin t cos t + ab (sin t)^3/cos t + ab (cos t)^3/sin t.
Ah, how nice and symmetrical. We must be doing something right.

OK, again I'm going to ignore the constant factor of ab, since it doesn't affect anything.
2 sin t cos t + (sin t)^3/cos t + (cos t)^3/sin t
And now for the big step. Differentiate:
[2 (cos t)^2 - 2 (sin t)^2] + [(3 (sin t)^2 (cos t)^2 + (sin t)^4)/(cos t)^2] + [(-3 (cos t)^2 (sin t)^2 - (cos t)^4)/(sin t)^2].

Put everything over a common denominator of (cos t)^2(sin t)^2.
OK, this is getting to hard to type, so I'm going to let A = cos t, B = sin t :)

The numerator becomes 2A^4B^2 - 2B^4A^2 + 3B^4A^2 + B^6 - 3A^4B^2 - A^6, which we want equal to 0.

We can simplify that slightly..
B^6 + B^4A^2 - B^2A^4 - A^6.
Now, divide by A^6, and let c=B/A.
c^6 + c^4 - c^2 - 1 = 0.
c^4(c^2 + 1) - (c^2 + 1) = 0
(c^4 - 1)(c^2 + 1) = 0
c^4 = 1 or c^2 = -1
The second case is impossible, so c^4 = 1, and c = 1 or -1.

Thus we need cos t / sin t = 1 or -1, so 1/tan t = 1 or -1, so tan t = 1 or -1.

That means t = 45 degrees or -45 degrees, but we want the one in the first quadrant, so t = 45 degrees.

Finally, we go back to start again:
x = a sqrt(2)/2, y = b sqrt(2)/2.

So y' = (-xb^2)/(ya^2) = -b/a.

So the equation is

y - b sqrt(2)/2 = (-b/a)(x - a sqrt(2)/2).

Or written nicely, ay + bx = absqrt(2).

Nice to see the answer comes to something simple.. theres probably a nicer proof :(

Answer 2

There is a much simpler way to do this that requires minimal math but considerable insight via a simple scaling transform. Make the substitutions: u = x/a, v = y/b

Now your ellipse becomes the unit circle, as follows:

b^2*x^2 + a^2*y^2 = a^2*b^2
Divide by a^2*b^2: (x/a)^2 + (y/b)^2 = 1
Substitute u and v: u^2 + v^2 = 1

Now find the equation of tangent line to the unit circle. A line with slope -m (all slopes are negative so m is positive) is tangent to the unit circle where the circle radius has a slope of 1/m. The coordinates of the point on the circle are determined by the two equations:

v/u = 1/m and u^2 + v^2 = 1

Combining: u^2 + (u/m)^2 = 1
Solving: u = sqrt(1/(1 + 1/m^2)) = sqrt(m^2/(m^2 + 1))
Solving for v:
since v^2 = 1 - u^2
v^2 = 1- m^2/(m^2 + 1) = 1/(m^2 + 1)
v = sqrt(1/(m^2 + 1))
Now you have an uv point on the tangent line and the slope (-m)
Using the point slope relation, all that remains is to find the intercept.

v = -mu + c
so c = v + mu = sqrt(1/(m^2 + 1)) + m*sqrt(m^2/(m^2 + 1))

This, being the v-intercept is the height of the triangle. The base of the triangle is the u-intercept for: 0 = -mu + c so u= c/m

The triangle height is c and base is c/m so the area is (c^2)/(2m)

Carry this out and after simplifying you get (c^2)/(2m) = (m^2 + 1)/2m

This is the area in scaled coordinates. We scaled x by a and y by b, that means that tha actual triangle is a time wider and b times higher than this one. Since ab is just a constant multiplier of the area, the maximum area here (in u,v space) represents a maximum area in xy space. So maximize the area as a function of m:
A = (m^2 + 1)/(2m)

dA/dm = 1 - (m^2 + 1)/(2m^2)

Set it equal to zero and solve for m:

0 = 1 - (m^2 + 1)/(2m^2)
2m^2 = m^2 + 1
m = 1

Since m = 1, c= sqrt(1/(m^2 + 1)) + m*sqrt(m^2/(m^2 + 1)) = 2*sqrt(1/2) = sqrt(2)

The triangle base is c/m so that is sqrt(2) as well so the area (in uv space) is sqrt(2)*sqrt(2)/2 = 1.

In xy space, the triangle is a times wider and b times higher so the minimum area is ab and the triangle dimensikons are multiplied by a nad b to be a*sqrt(2) wide by b*sqrt(2) high, the equation of the line follows directly since you have the two points (a*sqrt(2),0) and (0,b*sqrt(2)).

Again y = nx + d
b *sqrt(2) = n*0 + d so d = b*sqrt(2)
0 = n*a*sqrt(2) + b*sqrt(2) so n = -b/a

Giving a final equation of the line then is: y = (-b/a) * x + b*sqrt(2)

Answer 3

Solve equation of ellipse for y.
b^2*x^2 + a^2*y^2 = a^2*b^2
x^2/a^2 + y^2/b^2 = 1
y = b (1 - x^2/a^2)^(1/2)
y = (b/a) (a^2 - x^2)^(1/2)

Here's a handy substitution that I'll use now and then to clean up the algebra and the differentiation.
Q = (a^2 - x^2)^(1/2)

y = (b/a) Q

Find derivative, the slope of the tangent line.
dy/dx = -(b/a) x (a^2-x^2)^(-1/2)
dy/dx = -(b/a) x/Q

Form equation of tangent line.
Y = (dy/dx) X + Y0

Find y intercept of tangent line, in terms of x.
Y0 = Y - (dy/dx) X = y - (dy/dx) x
Y0 = (b/a) {(a^2 - x^2)^(1/2) + x^2 (a^2 - x^2)^(-1/2)}
Y0 = (b/a) {Q + x^2/Q}

Find x intercept, or root, of tangent line, in terms of x.
0 = (dy/dx) X0 + Y0
X0 = -Y0/(dy/dx)
X0 = -(b/a) {Q + x^2/Q} / {-(b/a) x/Q}
x0 = (1/x){Q^2 + x^2}
X0 = (1/x){a^2 - x^2 + x^2}
X0 = a^2/x

Form equation of the area of the triangle bounded by the tangent line and the coordinate axes, and solve for the area in terms of x.

A = (1/2) X0 Y0
A = (1/2) (a^2/x) (b/a) {(a^2 - x^2)^(1/2) + x^2 (a^2 - x^2)^(-1/2)}
A = (ab/2) (1/x) {(a^2 - x^2)^(1/2) + x^2 (a^2 - x^2)^(-1/2)}
A = (ab/2) {Q/x + x/Q}

Find the derivative of the triangle area with respect to the x coordinate of the tangent point.

dA/dx = (ab/2) (-1/x^2) {(a^2 - x^2)^(1/2) + x^2 (a^2 - x^2)^(-1/2)}
+ (ab/2) (1/x) { (1/2)(-2x)(a^2-x^2)^(-1/2) + 2x(a^2-x^2)^(-1/2) + x^2(-1/2)(-2x)(a^2-x^2)^(-3/2)}

dA/dx = - (ab/2) {(1/x^2)(a^2 - x^2)^(1/2) + (a^2 - x^2)^(-1/2)}
+ (ab/2) { (a^2-x^2)^(-1/2) + x^2(a^2-x^2)^(-3/2)}

dA/dx = (ab/2) {-(Q/x^2) - 1/Q + 1/Q + x^2/Q^3}

Find the triangle having minimum area by setting dA/dx=0 and solving for the coordinates of the tangent point [x,y] in terms of (a) and (b).

dA/dx = (ab/2) {x^2/Q^3 - Q/x^2} = 0

x^2/Q^3 = Q/x^2
Q = x = (a^2 - x^2)^(1/2)
x^2 = a^2 - x^2
2 x^2 = a^2
x^2 = a^2/2
x = a / sqrt(2)
y = (b/a) x
y = b / sqrt(2)

The y intercept of the tangent line is
Y0 = (b/a) {Q + x^2/Q}
Y0 = (b/a) 2x
Y0 = (b/a) 2a / sqrt(2)
Y0 = b sqrt(2)

The slope of the tangent line is
dy/dx = -(b/a) x/Q
dy/dx = -(b/a)

The equation of the tangent line is
Y = -(b/a) X + b sqrt(2)

The root of the tangent line is
X0 = a^2/x
X0 = a^2 / [a / sqrt(2)]
X0 = a sqrt(2)

The area of the triangle,
A = (ab/2) {Q/x + x/Q}
A = ab

Area of ellipse = pi a b
Ratio of ellipse area to triangle area = pi
Fraction of the triangle area that is outside the ellipse = 1 - pi/4

Answer 4

No College
start working earning money=no college debt=more job benefits=employer paying for college classes if you want=more money=success.

Answer 5

find the derivitive!

use the power rule and the chian rule and the multiplication rule all at once on each side, and simplify!

Done.