course Mth 271 ??????????????xassignment #012
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10:36:03 Class Notes #13 Explain how we obtain algebraically, starting from the difference quotient, the expression for the derivative of the y = x^2 function.
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RESPONSE --> y=2x
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10:36:25 ** The difference quotient is [ f(x + `dx ) - f(x) ] / `dx. In this case we get [ (x+`dx)^2 - x^2 ] / `dx = [ x^2 + 2 x `dx + `dx^2 - x^2 ] / `dx = [ 2 x `dx - `dx^2 ] / `dx = 2 x - `dx. Taking the limit as `dx -> 0 this gives us just 2 x. y ' = 2 x is the derivative of y = x^2. **
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RESPONSE --> I understand this
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10:37:24 **** Explain how the binomial formula is used to obtain the derivative of y = x^n.
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RESPONSE --> you would say y= x^n and the derivative would be y'= nx^(n-1)
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10:37:49 ** The key is that (x + `dx)^n = x^n + n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n. When we form the difference quotient the numerator is therefore f(x+`dx) - f(x) = (x + `dx)^n - x^n = (x^n + n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n) - x^n = n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n. The difference quotient therefore becomes (n x^(n-1)*`dx + n(n-1)/2 x^(n-2) * `dx^2 + ... + `dx^n) / `dx = n x^(n-1) + n (n-1) / 2 * x^(n-2) `dx + ... +`dx^(n-1). After the first term n x^(n-1) every term has some positive power of `dx as a factor. Therefore as `dx -> 0 these terms all disappear and the limiting result is n x^(n-1). **
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RESPONSE --> I am not doing this like you but I have the correct answer
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10:39:05 **** Explain how the derivative of y = x^3 is used in finding the equation of a tangent line to that graph.
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RESPONSE --> You would have to know x before you could obtain the tanget line....then you would use point slope and then you would get the tan line
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10:39:32 ** The derivative is the slope of the tangent line. If we know the value of x at which we want to find the tangent line then we can find the coordinates of the point of tangency. We evaluate the derivative to find the slope of the tangent line. Know the point and the slope we use the point-slope form to get the equation of the tangent line. **
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RESPONSE --> I think I got that.....you just explained it in better detail
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10:39:55 2.1.10 estimate slope of graph.................................................
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RESPONSE --> -1/3 .......to left 1 and up 3 units
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10:40:06 ** You can use any two points on the graph to estimate the slope. The slope of a straight line is the same no matter what two points you use. Of course estimates can vary; the common approach of moving over 1 unit and seeing how many units you go up is a good method when the scale of the graph makes it possible to accurately estimate the distances involved. The rise and the run should be big enough that you can obtain good estimates. One person's estimate: my estimate is -1/3. I obtained this by seeing that for every 3 units of run, the tangent line fell by one unit. Therefore rise/run = -1/3. **
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RESPONSE --> I understand that
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10:40:22 2.1.24 limit def to get y' for y = t^3+t^2
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RESPONSE --> y= 3t^2+ 2t
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10:40:29 ** f(t+`dt) = (t+'dt)^3+(t+'dt)^2. f(t) = t^3 + t^2. So [f(t+`dt) - f(t) ] / `dt = [ (t+'dt)^3+(t+'dt)^2 - (t^3 + t^2) ] / `dt. Expanding the square and the cube we get [t^3+3t^2'dt+3t('dt)^2+'dt^3]+[t^2+2t'dt+'dt^2] - (t^3 - t^2) } / `dt. } We have t^3 - t^3 and t^2 - t^2 in the numerator, so these terms subtract out, leaving [3t^2'dt+3t('dt)^2+'dt^3+2t'dt+'dt^2] / `dt. Dividing thru by `dt you are left with 3t^2+3t('dt)+'dt^2+2t+'dt. As `dt -> 0 you are left with just 3 t^2 + 2 t. **
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RESPONSE --> I understand
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10:42:31 2.1.32 tan line to y = x^2+2x+1 at (-3,4) What is the equation of your tangent line and how did you get it?
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RESPONSE --> 2x+2 f(-3)= 2(-3)+ (2) = -4 so slope = -4=m y-4= -4(x--3) y-4= -4x -12 y= -4x -8
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10:42:43 STUDENT SOLUTION WITH INSTRUCTOR COMMENT: f ' (x)=2x+2. I got this using the method of finding the derivative that we learned in the modeling projects The equation of the tangent line is 2x+2. I obtained this equation by using the differential equation. the slope is -4...i got it by plugging the given x value into the equation of the tan line. INSTRUCTOR COMMENT: If slope is -4 then the tangent line can't be y = 2x + 2--that line has slope 2. y = 2x + 2 is the derivative function. You evaluate it to find the slope of the tangent line at the given point. You have correctly found that the derivative is -4. Now the graph point is (-3,4) and the slope is -4. You need to find the line with those properties--just use point-slope form. You get y - 4 = -4(x - -3) or y = -4 x - 8. **
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RESPONSE --> I understant this one
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10:45:10 2.1.52 at what pts is y=x^2/(x^2-4) differentiable (graph shown) At what points is the function differentiable, and why?
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RESPONSE --> I thought that we were doing x^2/5 ........y'= 2/5 x^-3/5 ......I am not sure
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10:46:31 ** At x = -2 and x = +2 the function approaches a vertical asymptote. When the tangent line approaches or for an instant becomes vertical, the derivative cannot exist. The reason the derivative doesn't exist for this function this is that the function isn't even defined at x = +- 2. So there if x = +- 2 there is no f(2) to use when defining the derivative as lim{`dx -> 0} [ f(2+`dx) - f(2) ] / `dx. f(2+`dx) is fine, but f(2) just does not exist as a real number. **
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RESPONSE --> I see......2 and -2 are where the graph starts going into decemials and getting smaller
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10:46:56 **** Query 2.1.52 at what pts is y=x^2/(x^2-4) differentiable (graph shown)
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RESPONSE --> This is the same question?? ....... -2 and 2
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10:47:21 ** The derivative is defined on(-infinity,-2)u(-2,2)u(2,infinity). The reason the derivative doesn't exist at x = +-2 is that the function isn't even defined at x = +- 2. The derivative at 2, for example, is defined as lim{`dx -> 0} [ f(2+`dx) - f(2) ] / `dx. If f(2) is not defined then this expression is not defined. The derivative therefore does not exist. At x = -2 and x = +2 the function approaches a vertical asymptote. When the tangent line approaches or for an instant becomes vertical, the derivative cannot exist.**
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RESPONSE --> ok .....I should have put in the points....
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10:48:08 If x is close to but not equal to 2, what makes you think that the function is differentiable at x?
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RESPONSE --> because this is where the point gets smaller
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10:48:42 ** If x is close to 2 you have a nice smooth curve close to the corresponding point (x, f(x) ), so as long as `dx is small enough you can define the difference quotient and the limit will exist. **
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RESPONSE --> I should have said that there would have been a nice smooth curve
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10:49:50 If x is equal to 2, is the function differentiable? Explain why or why not.
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RESPONSE --> I am not sure....you would not have a nice smooth curve???
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10:50:13 GOOD ANSWER FROM STUDENT: if the function does not have limits at that point then it is not differentiableat at that point.
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RESPONSE --> I should have said then there would have been no lmits
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10:50:45 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I am not surew I understand about the last 2 problem about the nice smooth curve and about the no lim
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