mersenneforum.org can somebody plz help me with this calculus?
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 2005-09-14, 23:35 #1 JustinJS   8,039 Posts can somebody plz help me with this calculus? I'm having a little trouble with 2 problems because the day that they went over the problems i was out of school and my freakin teachers is never there after school to help me. 1) A particle moves along the parabola y = x^2 from the point (1,1) to the point (x,y) Show that (change in)y/(change in)x = x +1 2) Consider a circle of a radius 5, centered at (1,2). Find an equation of the line tangent to the circle at the point (-2, 6). Describe th procedure that you used to get your answer.
 2005-09-15, 00:13 #2 Ken_g6     Jan 2005 Caught in a sieve 2·197 Posts First, we don't do other peoples' homework. Second, I've only glanced over these problems. That said, I'll try to help... 1. This is easy! They've given you the answer; you just need to show work to find it. I suggest you find the change in each of (change in)x and (change in)y, and divide. If you don't get the answer they gave, you did it wrong! P.S. There's a chance you might have to do long division of polynomials (or factoring or something.) 2. It jumps out at me that the tangent point is not on either x=1 or y=2, yet the coordinates are integers. I'd draw a right triangle and see what you get.
2005-09-16, 00:21   #3
Bernard Schneid

Sep 2005

2 Posts
help with calculus

Quote:
 Originally Posted by JustinJS I'm having a little trouble with 2 problems because the day that they went over the problems i was out of school and my freakin teachers is never there after school to help me. 1) A particle moves along the parabola y = x^2 from the point (1,1) to the point (x,y) Show that (change in)y/(change in)x = x +1 2) Consider a circle of a radius 5, centered at (1,2). Find an equation of the line tangent to the circle at the point (-2, 6). Describe th procedure that you used to get your answer.
Looking at the second problem:

FIRST we need to determine the equation of the circle.

A circle with center at the origin and radius of 5 has the equation:

X^2 + Y^2 = 5^2=25

By analogy, your circle has a different center, so its equation is:

(X-1)^2 + (Y-2)^2 =25.

SECOND we need to find the slope of the circle at any point, which we get by taking the derivative implicitly:

Thus, the slope at any point is given by the following:

2(X-1) + 2 (Y-2) (dy/dx)=0

If we plug in the values X=-2 and Y= 6, we get

2(-3) + 2 (4) (dy/dx), we get that dy/dx=3/4.

To repeat, the slope of the tangent at that point is 3/4.

Since we know both the slope of the line, and a point
(-2,6) on the line, by the point/slope formula:

(Y-6)/(X+2)=3/4 ,and I leave the simplification to you.

 2005-09-16, 00:29 #4 Bernard Schneid   Sep 2005 2 Posts Looking at your first problem, we are drawing a STRAIGHT LINE between the point 1,1 and (y,x) The slope of the line is the change in y divided by the change in X, or : delta Y/ delta X. In other words: (Y-1)/(X-1). But Y=X^2, so that we can substitute, giving us: (X^2-1)/(X-1) But we can FACTOR the numerator, so that: (X-1) (X+1) / (X-1) cancelling (X-1) from the numerator and denominator, leaves us with the desired result: (X+1)
2005-09-16, 12:36   #5
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by Bernard Schneid Looking at the second problem: FIRST we need to determine the equation of the circle. A circle with center at the origin and radius of 5 has the equation: X^2 + Y^2 = 5^2=25 By analogy, your circle has a different center, so its equation is: (X-1)^2 + (Y-2)^2 =25. SECOND we need to find the slope of the circle at any point, which we get by taking the derivative implicitly: Thus, the slope at any point is given by the following: 2(X-1) + 2 (Y-2) (dy/dx)=0 If we plug in the values X=-2 and Y= 6, we get 2(-3) + 2 (4) (dy/dx), we get that dy/dx=3/4. To repeat, the slope of the tangent at that point is 3/4. Since we know both the slope of the line, and a point (-2,6) on the line, by the point/slope formula: (Y-6)/(X+2)=3/4 ,and I leave the simplification to you.

One can get the equation without any derivatives. The tangent line
will be othogonal to the radius at that point. The slope of the radiius
is (6-2)/(-2 - 1) = -4/3. Now, recall from elementary algebra that
if lline 1 has slope m1 and line 2 has slope m2, and they are othogonal,
that there is a very simple relation between m1 and m2 ..... (left as an

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