course Mth 271 ???v???]???????assignment #012
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14:42:28 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= x^2 Msec= f(x+dx) - f(x)/ dx =(x + dx)^2 - x^2/ dx = x^2 +2dx +(dx)^2 -x^2/ dx =2dx+dx^2/dx = 2 +dx as dx approaches 0 then = 2
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14:42:35
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17:03:10 ** 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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17:08:47 **** Explain how the binomial formula is used to obtain the derivative of y = x^n.
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RESPONSE --> the bionomial formula (x+a)^2= x^2+2ax+a^2 is used when inserting the dx in to the formula of f(x), (x + dx)^n or x^n +2dx + dx^2
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17:09:16 ** 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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17:11:14 **** 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 --> finding the derivative of y=x^3, y' = 3x^2 +/- c c= where tangent line crosses the y axis
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17:11:22 ** 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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17:11:29 2.1.10 estimate slope of graph.................................................
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17:11:36 ** 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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17:22:10 2.1.24 limit def to get y' for y = t^3+t^2
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RESPONSE --> y' = (t + dt)^3 +(t + dt)^2 - (t^3 + t^2) = 3t^2 + 6dt + 3dt^2 + t^2 + 2dt +dt^2 -t^3 -t^2 =
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17:22:59 ** 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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17:25:22 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 --> take derivative of the function gives the slope of the tangent line. y= x^2 + 2x +1 y'= 2x + 2 input value of x (-3 ) into equation y' = 4
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17:26:53 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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??w??????Y|??? assignment #012 ??????????D?Applied Calculus I 07-11-2006
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12:21:28 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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12:21:30
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12:21:34 ** 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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12:21:37 **** Explain how the binomial formula is used to obtain the derivative of y = x^n.
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12:21:39 ** 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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12:21:41 **** 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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12:21:43 ** 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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12:21:46 2.1.10 estimate slope of graph.................................................
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12:21:49 ** 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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12:21:52 2.1.24 limit def to get y' for y = t^3+t^2
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12:21:54 ** 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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12:22:04 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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12:22:53 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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12:25:00 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 --> -inf< 2< inf
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12:26:09 ** 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 --> ok
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12:27:38 **** Query 2.1.52 at what pts is y=x^2/(x^2-4) differentiable (graph shown)
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RESPONSE --> all points except 2 and -2
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12:28:04 ** 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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12:30:46 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 --> if x does not = +-2 then the denominator will not be 0. (division by 0 is undefined) if the denomintor is micro, nano, or pico we would still have a number other than 0
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12:30:57 ** 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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12:32:29 If x is equal to 2, is the function differentiable? Explain why or why not.
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RESPONSE --> no, the slope is 0, or the line is vertical. so there you cannot have a tangent line.
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12:32:36 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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12:36:50 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Professor, when the equation y=x^2 +2x +c was diff. and x point -3 was placed in equation so that we found -4. You said this was not the slope. But you used the point slope formula to further solve the equation coming up with y =-4x -8. can you explain a little further?
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g????r??????assignment #013 ??????????D?Applied Calculus I 07-15-2006
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07:03:03 2.2.20 der of 4 t^-1 + 1. Explain in detail how you used the rules of differentiation to obtain the derivative of the given function, and give your final result.
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07:05:19 ** STUDENT SOLUTION: To solve this using the rules of differentiation, I used the power and constant multiple rules. In dealing with t^-1, I applied the power rule and that gave me derivative -1t^-2. By the constant multiple we multiply this result by the constant 4 to get - 4 t^-2. To deal with 1, I used the constant rule which states that the derivative of a constant is 0. My final result was thus s'(t)=-4t^-2 + 0 = - 4 t^-2. **
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07:08:55 22.2.30 der of 3x(x^2-2/x) at (2,18) What is the derivative of the function at the given point?
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RESPONSE --> y=3x(x^2-2/x) y'= 3(2x +2x^-2) y'= 3[2(2)+2/2^2] y'=6
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07:11:08 ** You could use the product rule with f(x) = 3x and g(x) = x^2 - 2 / x. Since f ' = 3 and g ' = 2 x + 2 / x^2 we have (f g) ' = f ' g + f g ' = 3 (x^2 - 2 / x) + 3x ( 2x + 2 / x^2), which expands to (f g ) ' = 3 x^2 - 6 / x + 6 x^2 + 6 / x. This simplifies to give us just (f g ) ' = 9 x^2. It's easier, though, to just expand the original expression and take the derivative of the result: 3x ( x^2 - 2 / x ) = 3 x^3 - 6. The derivative, using the power-function rule, constant multiple rule and constant rule is thus y ' = 9x^2. At x = 2 we get derivative 9 * ( -2)^2 = 36. Note that (2, 18) is indeed on the graph because 3x ( x^2 - 2/x) evaluated at x = 2 gives us y = 3 * 2 ( 2^2 - 2 / 2) = 6 * 3 = 18. **
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RESPONSE --> I did not multiply the function before I took the derivative.
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07:14:15 Query 22.2.40 f'(x) for f(x) = (x^2+2x)(x+1) What is f'(x) and how did you get it?
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RESPONSE --> first multiply the function. y=(x^2+2x)(x+1) y= x^3+x^2+2x^2+2x y= x^3+3x^2+2x y'=3x^2+6x+2
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07:14:28 ** You could use the product rule, which would give you (x^2 + 2x) ' ( x + 1) + (x^2 + 2x) ( x + 1) ' = (2x + 2) ( x + 1) + (x^2 + 2 x ) ( 1) = 2 x^2 + 4 x + 2 + x^2 + 2 x = 3 x^2 + 6 x + 2. An easier alternative: If you multiply the expressions out you get x^3+3x^2+2x. Then applying the constant multiple rule and the simple power rule to the function you get f ' (x) = 3 x^2 + 6 x + 2 . **
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07:18:30 22.2.56 vbl cost 7.75/unit; fixed cost 500 What is the cost function, and what is its derivative?
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RESPONSE --> cost= 500(x+-7.75) cost'=500
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07:19:08 ** The terminology means that it costs 7.75 per unit to manufacture the item, and 500 to run the plant or whatever. So if you produce x units it's going to cost 7.75 * x, plus the 500. The cost function is therefore 7.75 x + 500. If you take the derivative of the cost function you are looking at the slope of a graph of cost vs. number produced. The rise between two points of this graph is the difference in cost and the run is the difference in the number produced. When you divide rise by run you are therefore getting the average change in cost, per unit produced, between those two points. That quantity is interpreted as the average cost per additional unit, which is the average variable cost. The derivative is the limiting value of the slope when you let the two graph points get closer and closer together, and so gives the instantaneous rate at which cost increases per additional unit. Note that the fixed cost doesn't influence this rate. Changing the fixed cost can raise or lower the graph but it can't change the slope. **
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RESPONSE --> i had the formula wrong.
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07:20:16 Why should the derivative of a cost function equal the variable cost?
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RESPONSE --> the variable cost represents the change in the cost just as a derivative represents the inst rate of change of a line or item
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07:20:26 ** The variable cost is defined as the rate at which the cost changes with repect to the number of units produced. That's the meaning of variable cost. That rate is therefore the derivative of the cost function. **
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??|??????????? assignment #014 ??????????D?Applied Calculus I 07-15-2006
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07:26:57 **** Query 2.3.8 ave rate compared with inst rates at endpts on [1,4] for x^-.5 **** What is the average rate of change over the interval and how did you get it?
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RESPONSE --> avg rate for the endpoints: y=f(x) y= x^5@ 1 =1 y=x^5@ 4 = 1024 (1024-1)/(4-1)= 341
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07:27:34 STUDENT SOLUTION: The average rate of change over the interval is -1/6. I got this answer by taking the difference of the numbers obtained when you plug both 1 and 4 into the function and then dividing that difference by the difference in 1 and 4. f(b)-f(a)/b-a
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RESPONSE --> not sure what i did wrong???
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07:29:13 **** How does the average compare to the instantaneous rates at the endpoints of the interval?
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RESPONSE --> inst= y'(1)= 5x^4 =5 y'(4)= 5x^4=1280
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07:30:10 ** The average rate of change is change in y / change in x. For x = 1 we have y = 1^-.5 = 1. For x = 4 we have y = 4^-.5 = 1 / (4^.5) = 1 / 2. So `dx = 1/2 - 1 = -1/2 and `dy = 4 - 1 = 3. `dy / `dx = (.5-1) / (4-1) = -.5 / 3 = -1/6 = -.166... . To find rates of change at endpoints we have to use the instantaneous rate of change: The instantaneous rate of change is given by the derivative function y ' = (x^-.5) ' = -.5 x^-1.5. The endpoints are x=1 and x=4. There is a rate of change at each endpoint. The rate of change at x = 1 is y ' = -.5 * 1^-1.5 = -.5. The rate of change at x = 4 is y ' = -.5 * 4^-1.5 = -.0625. The average of these two rates is (-.5 -.0625) / 2 = -.281 approx, which is not equal to the average rate -.166... . Your graph should show the curve for y = x^-.5 decreasing at a decreasing rate from (1, 1) to (4, .5). The slope at (1, 1) is -.5, the slope at (4, .5) is -.0625. and the average slope is -.166... . The average slope is greater than the left-hand slope and less than the right-hand slope. That is, the graph shows how the average slope between (1,1) and (4,.5), represented by the straight line segment between those points, lies between steeper negative slope at x=1 and the less steep slope at x = 4. If your graph does not clearly show all of these characteristics you should redraw the graph so that it does. **
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RESPONSE --> i was using 5 not 0.5
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07:40:01 Query 2.3.12 H = 33(10`sqrt(v) - v + 10.45): wind chill; find dH/dv, interpret; rod when v=2 and when v=5 What is dH/dt and what is its meaning?
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RESPONSE --> H=33(10sqrt(v)-v+10.45) H=330sqrt(v)-33v+344.85 dH/dv=165v^-1/2-33 v(2)=233.34-33= 200.34
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07:40:36 ERRONEOUS STUDENT SOLUTION: dH/dt is equal to [33(10 sqrt v+x - v+x + 10.45) + (33(10 sqrt v - v + 10/45))] / [(v+x)-x]. dH/dt represents the average heat loss from a person's body between two difference wind speeds; v+x and v INSTRUCTOR COMMENT: You give the difference quotient, which in the limit will equal the rate of change or the derivative, which is dH / dv = 33 * 10 * .5 * v^-.5 + 33 * -1 = 165 v^-.5 - 33. When v = 2, dH / dv is about 85 and when v = 5, dH / dv is about 40. Check my mental approximations to be sure I'm right (plug 2 and 5 into dH/dv = 165 v^-.5 - 33). H is the heat loss and v is the wind velocity. On a graph of H vs. v, the rise measures the change in heat loss and the run measures the change in wind velocity. So the slope measures change in heat loss / change in wind velocity, which is the change in heat loss per unit change in wind velocity. We call this the rate of change of heat loss with respect to wind velocity. dH / dv therefore measures the instantaneous rate of change of heat loss with respect to wind velocity. **
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