## Search found 21 matches

Wed Jul 01, 2009 9:08 pm
Forum: Calculus
Topic: For what values of x is the curve y=e^(-x^2) ...?
Replies: 2
Views: 2230

### Re: For what values of x is the curve y=e^(-x^2) ...?

$\frac{d^2y}{dx^2}=-2e^{-x^2}-2x\left(-2xe^{-x^2}\right)=-2e^{-x^2}+4x^2e^{-x^2}$
I did wonder if I'd messed up on the 2nd derivative. Thanks, Martingale!
Wed Jul 01, 2009 12:02 am
Forum: Calculus
Topic: For what values of x is the curve y=e^(-x^2) ...?
Replies: 2
Views: 2230

### For what values of x is the curve y=e^(-x^2) ...?

Problem : For what values of x is the curve y\,=\,e^{-x^2} concave downward? OK, so this calls for a 2nd derivative test. So this is what I did: \frac{dy}{dx}\,=\,-2xe^{-x^2} . That was easy. But now we have a product . \frac{d^2y}{dx^2}\,=\,-2[x(-2xe^{-x^2})\,-\,2e^{-x^2}]\,\,\,\,\,\,\,\,=\,4x^2e^...
Tue Jun 23, 2009 8:01 pm
Forum: Trigonometry
Topic: altitude 1208 m, angle of depr. 17 deg; find ground distance
Replies: 2
Views: 2919

### Re: altitude 1208 m, angle of depr. 17 deg; find ground distance

1) A ballonist records his altitude as 1208 meters. At the same time, his partner measures the angle of depression from their location to their landing spot to be 17 degrees. How far away, correct to the nearest meter, is the landing spot from a point on the ground directly below the balloon? Where...
Tue Jun 23, 2009 6:27 pm
Forum: Beginning Algebra
Topic: [MOVED] roots and the least common denominator
Replies: 2
Views: 2057

### Re: roots and the least common denominator

To find the least common denominator of {2 \over 3a}+{4 \over a^2} the book I'm reading says to multiply them to get a common denominator of 3a^2: {2a \over 3a^2}+{12 \over 3a^2} = {2a+12 \over 3a^2} But wouldn't it be more efficient to find the square roots of the second fraction before multiplyin...
Thu Jun 18, 2009 1:24 pm
Forum: Calculus
Topic: Required: y-coordinate of point(s) of tangency
Replies: 2
Views: 2278

### Re:

If you plug their solution in, you can see that it doesn't work. When x = 3, then the slope between the point on the curve and the origin is not equal to the derivative of the curve at that point. So the book's answer would appear to be in error. Too often we have found that to be the case. Growl. ...
Wed Jun 17, 2009 10:43 pm
Forum: Calculus
Topic: Required: y-coordinate of point(s) of tangency
Replies: 2
Views: 2278

### Required: y-coordinate of point(s) of tangency

Here is the problem: Find the y-coordinate of the points on the graph of y\,=\,ln\,\|x|^3 from which the tangents to the curve pass through the origin. Attempted solution : Drew the graph of y\,=\,ln\,\|x|^3 and sketched two lines, each containing the origin and intersecting the graph in one point. ...
Wed Jun 17, 2009 10:14 pm
Topic: Find an area of the triangle with the given vertices?
Replies: 2
Views: 6574

### Re: Find an area of the triangle with the given vertices?

I missed this part of class and so now cannot understand it at all. We're working with matrices right now, so it must have to do with those. Here are the directions: Find the area of the triangle with the given vertices. The first problem: A (0,1), B (2,6), C (5,5) If someone could step me through ...
Wed Jun 17, 2009 7:51 pm
Forum: Trigonometry
Topic: sine or cosine
Replies: 2
Views: 3475

### Re: sine or cosine

how do you know when to use sine law or cosine.
See next post below ...
Mon Jun 15, 2009 9:53 pm
Forum: Trigonometry
Topic: cos theta= -3/4, theta in quadrant 3. find sin theta, tan "
Replies: 1
Views: 15698

### Re: cos theta= -3/4, theta in quadrant 3. find sin theta, tan "

is it possible to have someone explain how to go about this in steps? please n thanks. Hmm... cos\,\theta\,=\,-\frac{3}{4},\,given\,\theta\,in\,Quadrant\,III Try this recipe: 1.) Draw a pair of coordinate axes, including labeling the axes x and y and the origin. A scale is not necessary. 2.) Draw t...
Thu Jun 11, 2009 11:52 pm
Forum: Beginning Algebra
Topic: solving system of equations: 3m + 2p = 1.8, 4m + 6p = 2.9
Replies: 2
Views: 2709

### Re: solving system of equations: 3m + 2p = 1.8, 4m + 6p = 2.9

3m+2p=1.80 4m+6p=2.90 solve for m and p Please help me solve this algebraically 3m\,+\,2p\,=\,1.80 4m\,+\,6p\,=\,2.90 Please review - Solving systems of equations by substitution : http://www.purplemath.com/modules/systlin4.htm Solving systems of equations by addition/elimination : http://www.purpl...