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Exercise 1
A tank holds 1000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show the volume V of water remaining in the tank (in gallons) after t minutes.
(a) If P is the point (15, 250) on the graph of V , find the slopes of the secant lines P Q when Q is the point on the graph with t = 5, 10, 20, 25, and 30.
(b) Estimate the slope of the tangent line at P by averaging the slopes of two secant lines. (c) Use a graph of the function to estimate the slope of the tangent line at P. (This slope represents the rate at which the water is flowing from the tank after 15 minutes.)
Solution
Below is a plot of the points listed in the table: (0, 1000), (5, 694), (10, 444), (15, 250), (20, 111), (25, 28), and (30, 0).
Part (a)
Use the slope formula for the two points on each secant line.
(15, 250) and (5, 694): m =
5 − 15 =^ −^44.^4
(15, 250) and (10, 444): m =^44410 −−^25015 = − 38. 8
(15, 250) and (20, 111): m =^111 −^250 20 − 15
(15, 250) and (25, 28): m =^2825 − −^250 15 = − 22. 2
(15, 250) and (30, 0): m =^0 −^250 30 − 15
The equations for the secant lines are obtained by using the point-slope formula.
V − 250 = − 44 .4(t − 15) V − 250 = − 38 .8(t − 15) V − 250 = − 27 .8(t − 15) V − 250 = − 22 .2(t − 15)
V − 250 ≈ − 16 .7(t − 15)
Plot these lines versus t with the scatterplot.
Exercise 3
The point P (2, −1) lies on the curve y = 1/(1 − x).
(a) If Q is the point (x, 1 /(1 − x)), use your calculator to find the slope of the secant line P Q (correct to six decimal places) for the following values of x:
(i) 1. 5 (ii) 1. 9 (iii) 1. 99 (iv) 1. 999 (v) 2. 5 (vi) 2. 1 (vii) 2. 01 (viii) 2. 001
(b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at P (2, −1). (c) Using the slope from part (b), find an equation of the tangent line to the curve at P (2, −1).
Solution
For each value of x, calculate y and then use the slope formula to find the slope of P Q.
x = 1. 5 → y = 1 1 − 1. 5
= −2 : Q(1. 5 , −2) ⇒ m = −^2 −^ (−1)
- 5 − 2
x = 1. 9 → y =
1 − 1. 9 =^ −^
9 :^ Q
⇒ m =
x = 1. 99 → y = 1 1 − 1. 99
: Q
⇒ m =
=^100
x = 1. 999 → y =
1 − 1. 999 =^ −^
999 :^ Q
⇒ m =
x = 2. 001 → y = 1 1 − 2. 001
: Q
⇒ m =
=^1000
x = 2. 01 → y =
1 − 2. 01 =^ −^
101 :^ Q
⇒ m =
x = 2. 1 → y = 1 1 − 2. 1
: Q
⇒ m =
=^10
x = 2. 5 → y =
1 − 2. 5 =^ −^
3 :^ Q
⇒ m =
The slope of the tangent line at x = 2 is presumably 1. Therefore, the equation of the tangent line is y + 1 = 1(x − 2),
or y =x − 3.
Exercise 7
The table shows the position of a motorcyclist after accelerating from rest.
(a) Find the average velocity for each time period: (i) [2, 4] (ii) [3, 4] (ii) [4, 5] (ii) [4, 6]
(b) Use the graph of s as a function of t to estimate the instantaneous velocity when t = 3.
Solution
Below is a plot of the motorcyclist’s position, using the points in the table: (0, 0), (1, 4 .9), (2, 20 .6), (3, 46 .5), (4, 79 .2), (5, 124 .8), and (6, 176 .7).
The average velocity in each time period is given by the slope of the line going through two points.
[2, 4] : m =
s(4) − s(2) 4 − 2 =
4 − 2 = 29.^3
ft s
[3, 4] : m = s(4)^ −^ s(3) 4 − 3
=^79.^2 −^46.^5
= 32. 7 ft s
[4, 5] : m = s(5) 5 −−^ s 4 (4) =^124. 58 −− 4 79.^2 = 45. 6 fts
[4, 6] : m =
s(6) − s(4) 6 − 4 =
6 − 4 = 48.^75
ft s
Exercise 9
The point P (1, 0) lies on the curve y = sin(10π/x).
(a) If Q is the point (x, sin(10π/x)), find the slope of the secant line P Q (correct to four decimal places) for x = 2, 1.5, 1.4, 1.3, 1.2, 1.1, 0.5, 0.6, 0.7, 0.8, and 0.9. Do the slopes appear to be approaching a limit? (b) Use a graph of the curve to explain why the slopes of the secant lines in part (a) are not close to the slope of the tangent line at P.
(c) By choosing appropriate secant lines, estimate the slope of the tangent line at P.
Solution
For each value of x, calculate y and then use the slope formula to find the slope of P Q.
x = 2 → y = 0 : Q(2, 0) ⇒ m =^0 −^0 2 − 1
x = 1. 5 → y ≈ 0 .866 : Q(1. 5 , 0 .866) ⇒ m =
1. 5 − 1 ≈^1.^73
x = 1. 4 → y ≈ − 0 .434 : Q(1. 4 , − 0 .434) ⇒ m = −^0.^434 −^0
- 4 − 1
x = 1. 3 → y ≈ − 0 .823 : Q(1. 3 , − 0 .823) ⇒ m =
1. 3 − 1 =^ −^2.^74
x = 1. 2 → y ≈ 0 .866 : Q(1. 2 , 0 .866) ⇒ m =^0.^866 −^0
- 2 − 1
x = 1. 1 → y ≈ − 0 .282 : Q(1. 1 , − 0 .282) ⇒ m =
1. 1 − 1 =^ −^2.^82
x = 0. 9 → y ≈ − 0 .342 : Q(0. 9 , − 0 .342) ⇒ m = −^00 ..^3429 −^ − 1 0 = 3. 42
x = 0. 8 → y = 1 : Q(0. 8 , 1) ⇒ m = 1 −^0
- 8 − 1
x = 0. 7 → y ≈ 0 .782 : Q(0. 7 , 0 .782) ⇒ m =
0. 7 − 1 =^ −^2.^61
x = 0. 6 → y ≈ 0 .866 : Q(0. 6 , 0 .866) ⇒ m =^0.^866 −^0
- 6 − 1
x = 0. 5 → y = 0 : Q(0. 5 , 0) ⇒ m =
The slopes of the secant lines do not appear to be approaching a limit.
This is because the increment of 0.1 is too large. Try to evaluate the secant line slope for the points at x = 1.01 and x = 1.001 instead.
x = 1. 01 → y ≈ − 0 .306 : Q(1. 01 , − 0 .306) ⇒ m = −^0.^306 −^0
- 01 − 1
x = 1. 001 → y ≈ − 0 .0314 : Q(1. 001 , − 0 .0314) ⇒ m = −^0.^0314 −^0
- 001 − 1
The slope of the tangent line at x = 1 is
dy dx
= cos(10π/x)
− 10 π x^2
dy dx
x=
= cos(10π)(− 10 π) ≈ − 31. 4159.
