Mountain Bike Reviews Forum banner
1 - 20 of 27 Posts

·
Registered
Joined
·
643 Posts
Discussion Starter · #7 ·
The Kadvang said:
take the derivative with implicit differentiation to find the slope, then plug the points in
well i found the distance from the focus of the parabola to the point of tnagency, but i dont know if i did it right.

i know that the distance from the focus to the point of tnagency is the same as the distance from the focus to the y-intercept of the tangent line but im not even sure i got the right distance
 

·
Registered
Joined
·
643 Posts
Discussion Starter · #8 ·
this cant be right...

i got the y-intercept to be inside of my parabola...

i used the equation b=p-d
p) being the focus of the parabola (according to my calculation its -1/8)
b) is the y-int
d) is the distance from focus to point of tange (according to my calc its 11/4)
 

·
yeah, uh............bikes
Joined
·
2,712 Posts
I'm not sure what method you're trying to use to find the line, but Kadvang was mostly correct.

Take the derivative and plug in the (x,y) for the point they gave you....this give you the slope of the tangent line. Then use the line equation y = m*x + b with the given point to find b.

MATH IS FUN
 

·
Registered
Joined
·
643 Posts
Discussion Starter · #12 ·
hmm...

FloridaFish said:
I'm not sure what method you're trying to use to find the line, but Kadvang was mostly correct.

Take the derivative and plug in the (x,y) for the point they gave you....this give you the slope of the tangent line. Then use the line equation y = m*x + b with the given point to find b.

MATH IS FUN
ya i still dont really get it, someone is gonna have to do the problem and like show me step by step cause im comin out with some weird stuff...
 

·
yeah, uh............bikes
Joined
·
2,712 Posts
ilikepinkbikes said:
ya i still dont really get it, someone is gonna have to do the problem and like show me step by step cause im comin out with some weird stuff...
plug (-2,-2) into the y = m*x +b where m is the slope found from the derivative equation. When b is found you have the equation for the line.
 

·
Registered
Joined
·
643 Posts
Discussion Starter · #14 ·
FloridaFish said:
plug (-2,-2) into the y = m*x +b where m is the slope found from the derivative equation. When b is found you have the equation for the line.
can you explain to me how to find the derivative equation because i dont think i have learned that yet. so these derivative terms are confusing me. Like i think i understand what you are doing but if i dont have another set of points i cant find the slope.
 

·
Registered
Joined
·
421 Posts
The first method is by the book. If you havn't covered implicit differentiation then follow the second example. The answers are the same.

1st method.

Take the derivative implicitly.

d/dx (x^2+2y)= 0
= d/dx (x^2) + d/dx (2y) = 0

Y is a function of X so use the chain rule.

d/dx (x^2) + d/dy (2y) dy/dx =0
= 2x + 2 dy/dx = 0

Solve for dy/dx. Remember that dy/dx is the slope of a line.

dy/dx = m = -x = -(-2)
m = 2

Use (y-y1) = m(x-x1) with (x1,y1) being your values. (point-slope equation)

(y2 - (-2) = 2(x2 - (-2)

solve for y
y = 2x + 2 (answer)

2nd method

x^2 + 2y = 0

solve for y
y = -x^2/2

take the derivative (which is the slope)

dy/dx (-x^2/2)
= -x
m = -x

The rest is the same as the first method.
Use (y-y1)=m(x-x1) with (x1,y1) being your values.

(y2 - (-2)) = 2(x2 - (-2))

solve for y
y= 2x + 2 (answer)
 

·
Renegade of Krunk
Joined
·
604 Posts
The Kadvang said:
take the derivative with implicit differentiation to find the slope, then plug the points in
Damn homie, you've taken Calc? I could be in pre-calc next semester (jr year) but seeing how I am totally sick of math and got a C in my last class I'm waiting till my senior year at the earliest.

Hooray for the Math section of AIM's tuesday. Not.
 

·
Registered
Joined
·
1,113 Posts
E30Evolution said:
Damn homie, you've taken Calc? I could be in pre-calc next semester (jr year) but seeing how I am totally sick of math and got a C in my last class I'm waiting till my senior year at the earliest.

Hooray for the Math section of AIM's tuesday. Not.
Yeah taking it this year as a junior, it sucks...teh AP is gonna destroy me.
 

·
Registered
Joined
·
643 Posts
Discussion Starter · #18 ·
muai wowy

jkish said:
The first method is by the book. If you havn't covered implicit differentiation then follow the second example. The answers are the same.

1st method.

Take the derivative implicitly.

d/dx (x^2+2y)= 0
= d/dx (x^2) + d/dx (2y) = 0

Y is a function of X so use the chain rule.

d/dx (x^2) + d/dy (2y) dy/dx =0
= 2x + 2 dy/dx = 0

Solve for dy/dx. Remember that dy/dx is the slope of a line.

dy/dx = m = -x = -(-2)
m = 2

Use (y-y1) = m(x-x1) with (x1,y1) being your values. (line/slope equation)

(y2 - (-2) = 2(x2 - (-2)

solve for y
y = 2x + 2 (answer)

2nd method

x^2 + 2y = 0

solve for y
y = -x^2/2

take the derivative (which is the slope)

dy/dx (-x^2/2)
= -x
m = -x

The rest is the same as the first method.
Use (y-y1)=m(x-x1) with (x1,y1) being your values.

(y2 - (-2)) = 2(x2 - (-2))

solve for y
y= 2x + 2 (answer)
i love you...
 

·
Renegade of Krunk
Joined
·
604 Posts
The Kadvang said:
Yeah taking it this year as a junior, it sucks...teh AP is gonna destroy me.
Yeah, I'm taking AP Bio next year cuz I really dont want to play around with AP Physics or Chem. Seeing as my teacher said like half the class fails, think I could find a better way to waste a class period a day and not get credit for it..Like study hall..Or no class for that matter. Good luck! My e-prayers are with you ;)
 

·
Registered
Joined
·
421 Posts
The antiderivative is a family of integrals with the form F(x) + C, where C is an arbitrary constant. For simple antiderivatives use the power rule. F(X)=x^(n+1)/(n+1) + C.

For example, the antiderivative of x^2 is X^3/3 + C. If you take the derivative of X^3/3 you get X^2, hence the name antiderivative.

For solving volume problems, remember that the integral is the area under the curve (infinite slices). (area x length = volume)
 
1 - 20 of 27 Posts
This is an older thread, you may not receive a response, and could be reviving an old thread. Please consider creating a new thread.
Top