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doin some calc at 2 a.m
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no...
no just calcjkish said:
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well.. im not doing calc homework.. but i'm a math major and way past calc if u need any help...
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i do need help.
x^2+2y=0; (-2,-2)
find the equation of the line tangent to the parabola at the given point.RockTheGeek said:
x^2+2y=0; (-2,-2)
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take the derivative with implicit differentiation to find the slope, then plug the points inilikepinkbikes said:
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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.The Kadvang said:
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
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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)
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)
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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
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
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hmm...
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...FloridaFish said:
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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.ilikepinkbikes said:
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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.FloridaFish said:
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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)
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)
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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.The Kadvang said:
Hooray for the Math section of AIM's tuesday. Not.
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Yeah taking it this year as a junior, it sucks...teh AP is gonna destroy me.E30Evolution said:
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muai wowy
i love you...jkish said:
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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 youThe Kadvang said:
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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)
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)
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