Topic: Who wants to help me with Calculus?

[ren]

I did most of it without any trouble but there's a few questions I'm stuck on. This is all in the chapter called Implicit differentiation. All I know how to do so far is find derivatives using chain rule, power rule, quotient rule and product rule.

4) At which point is the tangent to the curve x + y^2 = 1 parallel to the line x + 2y = 0? Answer: (0,1)
So for this question I've found the derivative of the tangent to be -1/2y and the slope as -2. Could be wrong though.

5b) Find two points on the ellipse at which the tangent is horizontal. Answer: (3√5/5,√5) (-3√5/5, -√5)
The equations 5x^2 - 6xy + 5y^2 = 16. I've already found that y' at point (1,-1) is equal to 1.

7) Find the equation of the normal at (2,3) to the curve x^3 + y^3 - 3xy = 17 at point (2,3). Answer: 7x - y - 11 = 0

I don't need all of them answered, but if someone could help me on one of them it'll probably point me in the right direction for the rest.

[Ghandi]

Wow. I got an A in calculus last year and I don't remember any of this. It's amazing how quickly I forget things.

[ren]

I almost got 7. I differentiated and got

3x^2 + 3y^2y' - 3y - 3xy' = 0. Then I subbed in the points (2,3) and found the slope of the tangent as -7 and ended up with 7x + y - 17 = 0. which is just slightly wrong.

[iPPi]

Haha I have no recollection of how to do this crap.

It's been 4 years since I've done any calculus like this. 

[ren]

I got 7 now. the slope of the tangent is -1/7, which means the slope of the normal is 7, which leads you to the right answer.

[DrSbaitso]

Because I was bored...

4) At which point is the tangent to the curve x + y^2 = 1 parallel to the line x + 2y = 0? Answer: (0,1)

-> x + 2y = 0 has slope -1/2
-> Differentiate curve with respect to x
1 + 2y*y' = 0
y' = -1/2y
-> Plug in y' = -1/2 to find point where slope of tangent is -1/2
y = 1
-> Solve for x in curve equation
x = 0

5b) Find two points on the ellipse at which the tangent is horizontal. Answer: (3√5/5,√5) (-3√5/5, -√5)
The equations 5x^2 - 6xy + 5y^2 = 16. I've already found that y' at point (1,-1) is equal to 1.

-> Differentiate ellipse equation with respect to x
10x - 6y - 6xy' + 10y*y' = 0
-> Plug in y' = 0
10x - 6y = 0
y = 5x/3
-> Plug this back into ellipse equation and solve
5x^2 - 6x(5x/3) + 5(5x/3)^2 = 16
(80/9)x^2 = 16
x^2 = 9/5
x = +- 3/sqrt(5)

7) Find the equation of the normal at (2,3) to the curve x^3 + y^3 - 3xy = 17 at point (2,3). Answer: 7x - y - 11 = 0

-> Differentiate curve w.r.t x
3x^2 + 3y^2*y' - 3y - 3xy' = 0
-> Plug in (2,3) and solve for y' (slope)
3*4 + 3*9*y' - 9 - 6y' = 0
y' = -3/21 = -1/7
-> Normal has slope negative reciprocal
m = 7

y = 7x + b
-> Plug in (2,3) and solve for b
3 = 14 + b
b = -11

-> So equation is
y = 7x - 11

[Ghandi]

I knew DrSbaitso would answer it. 

[Quemaqua]

I'm going to go hide behind my words now.  I have no fucking idea about anything anybody just said.

[ren]

Thanks a lot. I had already done 7 so I'm guessing you didn't read the thread, or you were just having fun.

Just some questions:

4)> x + 2y = 0 has slope -1/2
I thought slope in standard form was -b/a, which would just be -2.

Plug in y' = -1/2 to find point where slope of tangent is -1/2
how do I do that?
1+2yy' = 0
1+2y(-1/2)=0
1 - y = 0
y = -1

which would be wrong

5b) 5x^2 - 6x(5x/3) + 5(5x/3)^2 = 16
(80/9)x^2 = 16

how did you get to (80/9)x^2 = 16? that's more algebra than calc but I don't see it.

[ren]

wait I managed to get it to here, x = +- 3/sqrt(5). so that's x, and the y co-ordinate is 5, but from where...

[DrSbaitso]

4)> x + 2y = 0 has slope -1/2
I thought slope in standard form was -b/a, which would just be -2.

If you move it into y = mx + b form, you get
2y = -x
y = -x/2 = -(1/2)x

Plug in y' = -1/2 to find point where slope of tangent is -1/2
how do I do that?
1+2yy' = 0
1+2y(-1/2)=0
1 - y = 0
y = -1

It's all correct until the very last step. If 1 - y = 0, then y = 1, which is what the answer has.

5b) 5x^2 - 6x(5x/3) + 5(5x/3)^2 = 16
(80/9)x^2 = 16

how did you get to (80/9)x^2 = 16? that's more algebra than calc but I don't see it.

I did some simplifications in my head, may have made some mistakes but here goes :p
5x^2 - 6x(5x/3) + 5(5x/3)^2 = 16
5x^2 - 30x^2/3 + 5(25x^2/9) = 16
5x^2 - 10x^2 + (125/9)x^2 = 16
-5x^2 + (125/9)x^2 = 16
-(45/9)x^2 + (125/9)x^2 = 16
(80/9)x^2 = 16

x = +- 3/sqrt(5). so that's x, and the y co-ordinate is 5, but from where...

The point is on the ellipse, so you just substitute the value for x back into the ellipse equation to solve for y. Or, to make things easier, substitute back into y = 5x/3 since that will be tangent to the ellipse at x = +- 3/sqrt(5).

So using x = +3/sqrt(5) you get y = 5(3/sqrt(5))/3 = 5/sqrt(5) = sqrt(5)
and using x = -3/sqrt(5) you get y = 5(-3/sqrt(5))/3 = -5/sqrt(5) = -sqrt(5)

So the points are (3/sqrt(5), sqrt(5) and (-3/sqrt(5), -sqrt(5)).

[Cobra951]

*Sigh*  I took 12 hours of calculus at OSU, and those of you have have forgotten this in a few years may have some sympathy for the fact that it's mostly Greek to me now, after ~30 years.

[ren]

Thanks a lot. I actually enjoy Calculus quite a bit. Most of my problems stem from having bad algebra skills.

[Antares]

I had the same problem ren.  I didn't pay much attention in algebra and suffered because of it in calculus.  I never missed questions because I didn't know how to do them, but because I always f'ed up the algebra.  Oh well.

[scottws]

I was a calculus genius.  I was easily tops in my calc class in high school and it was my only A at Miami U. (of Ohio). 

However, "was" is the operative term there.  I don't remember any of it.  Completely Greek.  Though I'm sure I could pick it up again rather quickly if I was required to.