I'm not to good at these questions where you have to find the equation and I'm stuck with the one where you have to find the answer w/ 1st principles.
In case you can't see that pic:
http://img98.imageshack.us/img98/3199/810tz1.png
I'm not to good at these questions where you have to find the equation and I'm stuck with the one where you have to find the answer w/ 1st principles.
In case you can't see that pic:
http://img98.imageshack.us/img98/3199/810tz1.png
8. Remember that the derivative, dy/dx gives you the slope of the tangent line for any given value of x. So let's find dy/dx (or y', means the same thing), we will have to do this using implicit differentiation, i hope you know it.
x^3 + y^3 - 3xy = 17
=> 3x^2 + 3y^2 y' - 3y - 3x y' = 0 ....now we group the y' together
=> 3y^2 y' - 3x y' = -3x^2 + 3y
=> y'(3y^2 - 3x) = 3y -3x^2
=> y' = (3y - 3x^2)/(3y^2 - 3x)
now at (2,3), y' = [3(3) - 3(2)^2]/[3(3)^2 - 3(2)] = (-3)/(21) = -1/7
now recall that the equation of a line is of the form y = mx + b, where m is the slope given by the derivative and b is the y-intercept. we will use the point-slope form to find the equation of the tangent line.
y - y1 = m(x - x1)
=> y - 3 = (-1/7)(x - 2)
=> y = (-1/7)x + 2/7 + 3
=> y = (-1/7)x + 23/7 ............equation of the tangent line.
9. Given the curve y = kx^2 + 5 and the eqution of the tangent line 4x + y = 9
First let's see where these lines intersect. we have:
y = -4x + 9
y = kx^2 + 5
=> kx^2 + 5 = -4x + 9
=> kx^2 + 4x - 4 = 0
=> x = [-4 +/- sqrt(16 + 16k)]/2k ............by the quadratic formula
=> x = [-4 +/- -4sqrt(1 + k)]/2k
=> x = [-2 +/- -2sqrt(1 + k)]/k
now y = kx^2 + 5
=> y' = 2kx
since the slope of the given tangent line is -4, we want y' = 2kx = -4
=> 2kx = -4
=> kx = -2
=> x = -2/k
now substitute this value for x in the formula we made above.
x = [-2 +/- -2sqrt(1 + k)]/k
=> -2/k = [-2 +/- -2sqrt(1 + k)]/k
=> -2 = -2 +/- -2sqrt(1 + k)
=> 0 = +/- -2sqrt(1 + k)
=> 0 = +/- sqrt(1 + k)
=> 0 = 1 + k ................squared both sides
=> k = -1
10. f(x) = sqrt(2 - x)
they want us to find the derivative of f(x) using first principles. this means they want us to find it by the limit definition:
using f ' (x) = limit{h-->0} [f(x + h) - f(x)]/h
=> f ' (x) = limit{h-->0} [sqrt(2 - x - h) - sqrt(2 - x)]/h ...now we abandon our mathematical training for a moment and irrationalize the quotient.
=> f ' (x) = limit{h-->0} [sqrt(2 - x - h) + sqrt(2 - x)]/[sqrt(2 - x - h) + sqrt(2 - x)] * [sqrt(2 - x - h) - sqrt(2 - x)]/h
=> f ' (x) = limit{h-->0} [2 - h - x - (2 - x)]/h[sqrt(2 - x - h) + sqrt(2 - x)]
=> f ' (x) = limit{h-->0} -h/h[sqrt(2 - x - h) + sqrt(2 - x)]
=> f ' (x) = limit{h-->0} -1/[sqrt(2 - x - h) + sqrt(2 - x)]
finally we can take the limit without worrying about the denominator going to zero.
=> f ' (x) = -1/[sqrt(2 - x) + sqrt(2 - x)]
=> f ' (x) = -1/[2*sqrt(2 - x)]
=> f ' (x) = -(1/2)(2 - x)^(-1/2) ..........as expected
if there's anything you don't understand about any of the questions, dont hesitate to ask
I don't understand some parts of the second one (#9)
When you go from
x = [-4 +/- sqrt(16 + 16k)]/2k ............by the quadratic formula
to
=> x = [-4 +/- -4sqrt(1 + k)]/2k
How do you take the (-4) out and put it in front of the square root?
Also, when you go from 0 = +/- -2sqrt(1 + k)
to:
=> 0 = +/- sqrt(1 + k)
Do you divide both sides by (-2) to get rid of the -2?
x = [-4 +/- sqrt(16 + 16k)]/2k
x = [-4 +/- sqrt[16(1+k)] ]/2k
x = [-4 +/- sqrt(16).sqrt(1+k)]/2k
and you know sqrt(16)= +/- 4
taking either works, since we already have a +/- it would just change it to a -/+
x = [-4 +/- -4sqrt(1+k)]/2k
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to your second question, yes, just divide both side by -2.
Sorry to bother you again, but I just can't understand the one step in the third question (#10)
How do you go from this:
=> f ' (x) = limit{h-->0} [sqrt(2 - x - h) - sqrt(2 - x)]/h
to this:
f ' (x) = limit{h-->0} [sqrt(2 - x - h) + sqrt(2 - x)]/[sqrt(2 - x - h) + sqrt(2 - x)] * [sqrt(2 - x - h) - sqrt(2 - x)]/h
Edit: Never mind, I see what you did. I wrote it out again on paper without the sqrt or ^ written and saw it. Thanks anyways :-)