Help!!!! Calculus!?

Question

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)>0.

Answer 1

This is a very nice challenging problem. You are lucky to have such a good teacher challenge you like this.

First, use implicit differentiation to find dy/dx. Remember that a and b are constants:

2 b^2 x dx+2 a^2 y dy=0

2 a^2 y dy=-2 b^2 x dx

dy/dx = -b^2/a^2 x/y

Now let (c,d) be any point on the ellipse that is in the first quadrant.

An equation of the tangent to the ellipse in point slope form is

y-d=-b^2/a^2 c/d (x-c)

The x-intercept of this equation is found by setting y=0:

(c+(a^2d^2)/(b^2 c),0)

and the y-intercept of this equation is found by setting x=0:

(0,d+(b^2 c^2)/(a^2 d))

and the area of the triangle formed by these two points and the origin is

1/2 * (d+(b^2 c^2)/(a^2 d)) * (c+(a^2d^2)/(b^2 c))=
(b^2 c^2+a^2 d^2)^2/(2a^2b^2cd)

Solving

b^2*c^2 + a^2*d^2 = a^2*b^2

for d gives us

d=(b*Sqrt[a^2 - c^2])/a

and substituting this value into

(b^2 c^2+a^2 d^2)^2/(2a^2b^2cd)

gives us the following function of c:

f(c)=(a^3*b)/(2*c*Sqrt[a^2 - c^2])

DIfferentiating with respect to c gives us

f'(c)=-(a^3*b*(a^2 - 2*c^2))/(2*c^2*(a^2 - c^2)^(3/2))

and solving f'(c)=0 results in c=a/Sqrt[2].

Challenge: Repeat the problem for the parabola y=a^2-b^2x^2, where a and b are positive.

Good luck! :)

Answer 2

An existing answer does this with implicit differentiation, which reduces workload slightly because it does not require putting everything in terms of x. If you feel compelled to put everything in terms of x, try this answer.

Rewrite your ellipse as:

y^2 = b^2 - (b/a)^2 x^2

If you're interested in quadrant 1, then all you care about is the positive square root of y. This forms a function:

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

If you take the derivative of that function with respect to x, you'll get another function y'(x) that will give you the slope of the line tangent to the curve at any point x. (and, for quadrant 1, we are restricting our x to x>0) I'll let you take the derivative, but your result will be:

y'(x) = -(b/a)^2 x/y

As another answer has shown, this is the result you get from implicit differentiation. Your answer will match this if you replace y with y(x) above.

Now we need to find the tangent line that gives us the TRIANGLE OF GREATEST AREA. In other words, we need a formula:

0.5*H*W

where H is the y-intercept of the tangent line and W is the x-intercept of the tangent line. We need this for every x0 under the ellipse and then need to maximize that function. Before we can do any of that, we need to get expressions for H and W. Before we can do that, we need an expression for every tangent line formed at an arbitrary point ( x0, y(x0) ). The expression for such a line is:

y - y(x0) = y'(x0) * (x - x0)

Which in slope-intercept form is:

y = y'(x0)*x + y(x0) - x0*y'(x0)

It's clear that the y-intercept (set x=0 and solve for y) for this line is:

H = y(x0) - x0*y'(x0)

Now we just need the x-intercept. If you set y=0 and solve for x in the equation for the line, you then get:

W = ( x0*y'(x0) - y(x0) )/ y'(x0)

Now you have your formula for the area of the triangle:

0.5*H*W = 0.5*( y(x0) - x0*y'(x0) )*( x0*y'(x0) - y(x0) )/ y'(x0)

Note that that's equal to:

-0.5*( y(x0) - x0*y'(x0) )^2/ y'(x0)

This negative sign makes sense because y'(x0) will be negative and so this will be a positive area.

Now all you have to do is maximize this function by taking the derivative with respect to x0. Find the point x0 that makes the derivative 0 (and second derivative negative). Once you have that, plug in x0 and y(x0) into your equation for the line:

y = y'(x0)*x + y(x0) - x0*y'(x0)

And that's your answer.

All I've left you do to is derive y'(x0) and do the maximization problem.

Answer 3

"I'll assume the x-axis is the axis of rotation.
posted by fasteddie1965 on 3/15/05 1:36 PM
(b^2*x^2)+(a^2*y^2)=a^2*b^2.
x^2/a^2 + y^2/b^2 = 1
y^2/b^2 = 1 - x^2/a^2
y^2 = b^2(1 - x^2/a^2)
y = +/- b sqrt(1 - x^2/a^2)


Using the disk method:
dV = 2xydx = 2xb sqrt(1 - x^2/a^2) dx
(1/2)V = int [2xb sqrt(1 - x^2/a^2) dx] from x = 0 to x = a.
V = 2 int [2xb sqrt(1 - x^2/a^2) dx] from x = 0 to x = a.
u = 1 - x^2/a^2
du = -2x/a^2 dx
-a^2 du = 2x dx
V = -a^2 [2b u^(1/2) du] from u = 1 to u = 0.
V = -a^2 (2)b (2/3) u^(3/2)
Evaluating:
-a^2 (4/3)b (1 - 0) = (4/3) a^2b"

Answer 4

hahahhaaaaa help yourself!!!!

Answer 5

how would i know?????