qa 14

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19:44:56 `q001. What are the coordinates of the x = 5 point and the slope at that point of the graph of the function y = .3 x^2? What is the equation of the line through the point and having that slope? Sketch the line and the curve between x = 3 and x = 7 and describe your sketch. How close, in the vertical direction, is the line to the graph of the y = .3 x^2 function when x = 5.5, 6, 6.5 and 7?

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RESPONSE --> y = .3 x^2 y = .3 (5)^2 y=7.5 so (5, 7.5) y = .3 x^2 so y' = (2).3x^1 or y'=.6x 'y(5)=3=slope y=3x+b 7.5=3(5)+b b= -7.5 equation for that line is: y=3x-7.5 it looks kinda like an upside down rainbow touching the ground if you turn it right. y=3(5.5)-7.5 (5.5,9) y = .3 (5.5)^2 (5.5,9.075) they are vertically.075 apart y=3(6)-7.5 (6,10.5) y = .3 (6)^2 (6,10.8) they are vertically.8 apart

The very last result, the .8 difference, differs a bit from the given solution but everything else is great. Good work on this question.

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19:47:15 The x = 5 point has y coordinate y = .3 * 5^2 = 7.5, so the point lies at (5, 7.5). The slope that that point is found by evaluating the derivative at that point. The derivative is y ' = .6 x, so the slope is y ' = .6 * 5 = 3. The line therefore passes through the point (5,7.5) and has slope 3. Its equation is therefore y - 7.5 = 3 ( x - 5), which simplifies to y = 3 x - 7.5. When x = 5.5, 6, 6.5 and 7, the y coordinates of the line are y = 3 ( 5.5 ) - 7.5 = 9, y = 3 (6) -7.5 = 10.5, y = 3 (6.5) -7.5 = 12, y = 3 (7) -7.5 = 13.5. At the same x coordinates the function y =.3 x ^ 2 takes values y = .3 (5.5) ^ 2 = 9.075, y =.3 (6) ^ 2 = 10.8, y =.3 (6.5) ^ 2 = 12.675, and y =.3 (7) ^ 2 = 14.7. The straight line is therefore lower than the curve by 9.075 - 9 = .075 units when x = 5.5, 10.8-10.5 = .3 when x = 6, 12.675-12 = .675 when x = 6.5, and 14.7-13.5 = 1.2 when x = 7. We see that as we move away from the common point (5,7.5), the line moves away from the curve more and more rapidly. Your sketch should show a straight line tangent to the parabolic curve y =.3 x^2, with the curve always above the line except that the point tangency, and moving more and more rapidly away from the line the further is removed from the point (5,7.5). The line you have drawn is called the line tangent to the curve at the point (5,7.5).

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RESPONSE --> ok

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20:02:06 `q002. What is the equation of the line tangent to the curve y = 120 e^(-.02 t) at the t = 40 point? If we follow the tangent line instead of the curve, what will be the y coordinate at t = 40.3? How close will this be to the y value predicted by the original function? Will the tangent line be closer than this or further from the original function at t = 41.2?

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RESPONSE --> y = 120 e^(-.02 t) der. is the same. y' = 120 e^(-.02 (40)) y'(40)=267.06 y=mx+b 267.06= 267.06(40)+b b= -10415.34 is this right? i think i messed up.

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20:06:34 The tangent line at the x = 40 point will have y coordinate y = 120 e^(-.02 * 40) = 53.9194, accurate to six significant figures. The derivative of the function is y ' = -2.4 e^(-.02 t) and at x = 40 has value y ' = -2.4 e^(-.02 * 40) = -1.07838, accurate to six significant figures. The tangent line will therefore pass through the point (40, 53.9194) and will have slope 1.07838. Its equation will therefore be y - 53.9194 = -1.07838 ( x - 40), which we can solve for y to obtain y = -1.07838 x + 97.0546. The values of the function at x = 40.3 and x = 41.2 are 53.5969 and 52.6408. The y values corresponding to the tangent line are 53.5958 and 52.6253. We see that the tangent line at x = 40.3 is .0011 units lower than the curve of the function, and at x = 41.2 the tangent line is .0155 units lower--about 14 times as far from the curve as at the x = 40.3 point.

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RESPONSE --> ok. i think i didn't use the (-) ?

That would do it. Big difference. Otherwise your approach was great.

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20:14:01 `q003. Sketch a graph of this curve and the t = 40 tangent line and describe how the closeness of the tangent line to the curve changes as we move away from the t = 40 point.

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RESPONSE --> i would assume it gets farther away?

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20:17:07 Your sketch should show the gradually curving exponential function very close to the tangent line. The tangent line is straight while the slope of the exponential function is always increasing. At the t - 40 point the two meet for an instant, but as we move both to the right and to the left the exponential curve moves away from the

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RESPONSE --> ok. (my derive program expired a while ago and it is a pain in the ass to graph that on the ti92- set limits, combine functions, zoom in/out/box like 30 times, etc.)

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20:17:21 tangent line, moving away very gradually at first and then with increasing rapidity.

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RESPONSE --> o...k.....

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20:32:00 `q004. What is the equation of the line tangent to the curve y = 20 ln( 5 x ) at the x = 90 point?

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RESPONSE --> y = 20 ln( 5x) y'=20/5x?? y'=20/5(90) y'(90)=.0444 y = 20 ln( 5(90)) (90,122.185) (122.185)=.044(90)+b b=118.225 y=.044x+118.225

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20:32:42 At x = 90, we have y = 20 ln (5 * 90) = 20 ln(450) = 20 * 6.1 = 126, approx.. So we are looking for a straight line through the point (90, 126). The slope of this straight line should be equal to the derivative of y = 20 ln(5 x) at x = 90. The derivative of ln(5x) is, by the chain rule, (5x) ' * 1 / (5x) = 5 / (5x) = 1/x. So y ' = 20 * 1 / x = 20 / x. At x = 90 then we have y ' = 20 / 90 = .22, approx.. Thus the tangent line passes through (90, 126) and has slope .22. The equation of this line is ( y - 126 ) / (x - 90) = .22, which is easily rearranged to the form y = .22 x + 106.

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RESPONSE --> y=.044x+118.225 um....help?

Your method is again fine. However y ' = 20 / x, not 20 / (5x) -- you neglected to use the chain rule so you got a slope of .044 rather than .22.

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`q004. What is the equation of the line tangent to the curve y = 20 ln( 5 x ) at the x = 90 point?

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RESPONSE --> ok so the ^5x is another function. i don't get it.

ln(5x) is a composite f(g(x)) with f(z) = ln(z) and g(x) = 5x. You apply the chain rule. f ' (z) = 1 / z, g ' (x) = 5 so ( f(g(x) ) ' = g'(x) * f ' (g(x)) = 5 * 1 / (5x) = 1 / x. Multiplying by 20, you get y ' = 20 * 1 / x = 20 / x.

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19:47:43 At x = 90, we have y = 20 ln (5 * 90) = 20 ln(450) = 20 * 6.1 = 126, approx.. So we are looking for a straight line through the point (90, 126). The slope of this straight line should be equal to the derivative of y = 20 ln(5 x) at x = 90. The derivative of ln(5x) is, by the chain rule, (5x) ' * 1 / (5x) = 5 / (5x) = 1/x. So y ' = 20 * 1 / x = 20 / x. At x = 90 then we have y ' = 20 / 90 = .22, approx.. Thus the tangent line passes through (90, 126) and has slope .22. The equation of this line is ( y - 126 ) / (x - 90) = .22, which is easily rearranged to the form y = .22 x + 106.

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RESPONSE --> ???

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19:59:29 `q005. Using the tangent line and the value of the function at x = 3, estimate the value of f(x) = x^5 and x = 3.1. Compare with the actual value.

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RESPONSE --> f(x) = x^5 f(x)' =5x^4 f(x)' =5(3)^4 f(5)' =405 wait. am i doing this right? i'm confused...

f(x)' =5x^4. That's it. No further steps necessary, except of course to plug in x = 5.

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20:10:45 f ' (x) = 5 x^4. When x = 3, f(x) = 3^5 = 243 and f ' (x) = 5 * 3^4 = 405. Thus tangent line passes through (3, 243) and has slope 405. Its equation is therefore (y - 243) / (x - 3) = 405, which simplifies to y = 405 x - 972. This function could be evaluated at x = 3.1. However a more direct approach simply uses the differential. In this approach we note that at x = 3 the derivative, which gives the slope of the tangent line, is 405. This means that between (3, 243) and the x = 3.1 point the rise/run ratio should be close to 405. The run from x = 3 to x = 3.1 is 1, so a slope of 405 therefore implies a rise equal to slope * run = 40.5. A rise of about 40.5 means that y = 3.1^5 should change by about 40.5, from 243 to about 283.5. The actual value of 3.1^5 is about 286.3. The discrepancy between the straight-line approximation and the actual value is due to the fact that y = x^5 is concave up, meaning that the slope is actually increasing. The closely-related idea of the differential, which will be developed more fully in the next section, is that if we know the rate at which the function is changing at a given point, we can use this rate to estimate its change as we move to nearby points.

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RESPONSE --> o...k...

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20:26:47 `q006. Using the differential and the value of the function at x = e, estimate the value of ln(2.8). Compare with the actual value.

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RESPONSE --> ok. i'm lost. i finally get the way you used the decimals above.but ln(2.8) x = e???? help?

e is approximately 2.71828. You use the function f(x) = ln(x), and do a tangent-line approximation at the point where x = e. ln(e) = 1 because the natural log is the inverse of the exponential function. So you are using the point (1, e) as the known point of the straight line. f ' (x) = 1 / x, so at x = e the slope is of that line is 1 / e.