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 January 30th, 2018, 11:19 PM #1 Senior Member   Joined: Apr 2008 Posts: 193 Thanks: 3 What am I missing in this calculus problem? shadow.pdf Please see the diagram attached to this post. (ex) The figure shows a lamp located 3 units to the right of the y-axis and a shadow created by the elliptical region x^2 + 4y^2 < = 5. If the point (-5,0) is on the edge of the shadow, how far above the x-axis is the lamp located? my attempt let (x,y) be the position of the light bulb. Next, set up the distance formula from (-5,0) to (x,y). D=sqrt((x+5)^2+y^2) Change the given inequality to equality and then isolate y^2. I get y^2=(5-x^2)/4 and substitute it into the distance formula. Square the formula and then let f(x)=D^2. Then, I have f(x)=(x+5)^2 + (5-x^2)/4. Next, differentiating the function, setting it to zero and solving for x give the critical value x=-20/3. Lastly, plug this value of x into y^2=(5-x^2)/4. But, y^2=(5-400/9)/4 < 0, which is impossible. I think I am missing something here. Could someone please explain how to do this problem? Thank you very much. Last edited by skipjack; January 31st, 2018 at 08:55 AM.
 January 31st, 2018, 03:18 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra The edge of the shadow is a straight line that is tangent to the ellipse. Thus we find an equation for the ellipse and from that, the slope of the tangent at any point $A=(x_0,y_0)$ on the ellipse. This is then also the slopes of the tangent at that point, and thus also the slope of the edge of the shadow. You can then use the point of intersection $A$ and the slope of the tangent to find the equation of the tangent through $A$ ($y=mx+c$ where $m$ and $c$ depend on $x_0$ and $y_0$). By eliminating $y_0$ from the equation and setting $(x,y)=(-5,0)$, you determine a value for $x_0$. This fixes exactly the equation of the tangent in which you set $x=3$ to find $y$, the height of the light source. Thanks from davedave
January 31st, 2018, 03:51 AM   #3
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Quote:
 Originally Posted by davedave Attachment 9403 Please see the diagram attached to this post. (ex) The figure shows a lamp located 3 units to the right of the y-axis and a shadow created by the elliptical region x^2 + 4y^2 < = 5. If the point (-5,0) is on the edge of the shadow, how far above the x-axis is the lamp located? my attempt let (x,y) be the position of the light bulb.
You are given that x = 3, so this is (3, y).

Quote:
 Next, set up the distance formula from (-5,0) to (x,y).
Why? The distance from the light to the shadow is not relevant. Rather, you should be using the equation of the line from (-5, 0). Since I want to use x and y as variables, call the position of the light (3, b). The equation of the line through (3, b) and (-5, 0) is y = (b/8)(x + 5). Now, for what value of b is that tangent to the ellipse x^2 + 4y^2 = 5? At a point where they intersect, they have the same (x, y) values, so we can substitute y = (b/8)(x + 5) for y in the equation of the ellipse: x^2 + 4(b/8)^2(x + 5)^2 = x^2 + (b^2/16)x^2 + (5b^2/8)x + (25b^2/16) = 5.

((16 + b^2)/16)x^2 + (5b^2/8)x + (25b^2 + 80)/5 = 0

That's a quadratic equation. It will have a single root when the left side is a perfect square.

Quote:
 Originally Posted by davedave D=sqrt((x+5)^2+y^2) Change the given inequality to equality and then isolate y^2. I get y^2=(5-x^2)/4 and substitute it into the distance formula. Square the formula and then let f(x)=D^2. Then, I have f(x)=(x+5)^2 + (5-x^2)/4. Next, differentiating the function, setting it to zero and solving for x give the critical value x=-20/3. Lastly, plug this value of x into y^2=(5-x^2)/4. But, y^2=(5-400/9)/4 < 0, which is impossible. I think I am missing something here. Could someone please explain how to do this problem? Thank you very much.

Last edited by skipjack; January 31st, 2018 at 08:59 AM.

 February 5th, 2018, 10:47 PM #4 Senior Member   Joined: Apr 2008 Posts: 193 Thanks: 3 Thank you, v8archie and Country Boy.

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