now, let's recall Cavalieri's Principle, "solids with equal altitudes and cross-sectional areas at each height, have the same volume".
so, even though this cylinder is slanted, it has the same cross-sectional area as a cylinder that's straight-up, whose altitude is h = 5, and radius r = 4.
Answer:
(a) The coordinates of the points of intersection are (-2, -12/5), (2/5, 0), and (2, 8/5)
(b) The gradient of the curve at each point of intersection are;
Gradient at (-2, -12/5) = -0.92
Gradient at (2/5, 0) = 4.3
Gradient at (2, 8/5) = -0.28
Step-by-step explanation:
The equations of the lines are;
5·x - 5·y = 2......(1)
x²·y - 5·x + y + 2 = 0.......(2)
Making y the subject of equation (1) gives;
5·y = 5·x - 2
y = (5·x - 2)/5
Making y the subject of equation (2) gives;
y·(x² + 1) - 5·x + 2 = 0
y = (5·x - 2)/(x² + 1)
Therefore, at the point the two lines intersect their coordinates are equal thus we have;
y = (5·x - 2)/5 = y = (5·x - 2)/(x² + 1)
Which gives;
Therefore, 5 = x² + 1
x² = 5 - 1 = 4
x = √4 = 2
Which is an indication that the x-coordinate is equal to 2
The y-coordinate is therefore;
y = (5·x - 2)/5 = (5 × 2 - 2)/5 = 8/5
The coordinates of the points of intersection = (2, 8/5}
Cross multiplying the following equation
Substituting the value for y in equation (2) with (5·x - 2)/5 gives;
Therefore;
5·x³ - 2·x² - 20·x + 8 = 0
(x - 2)×(5·x² - b·x + c) = 5·x³ - 2·x² - 20·x + 8
Therefore, we have;
x²·b - 2·x·b -x·c + 2·c -5·x³ + 10·x²
5·x³ - 10·x² - x²·b + 2·x·b + x·c - 2·c = 5·x³ - 2·x² - 20·x + 8
∴ c = 8/(-2) = -4
2·b + c = - 20
b = -16/2 = -8
Therefore;
(x - 2)×(5·x² - b·x + c) = (x - 2)×(5·x² + 8·x - 4)
(x - 2)×(5·x² + 8·x - 4) = 0
5·x² + 8·x - 4 = 0
x² + 8/5·x - 4/5 = 0
(x + 4/5)² - (4/5)² - 4/5 = 0
(x + 4/5)² = 36/25
x + 4/5 = ±6/5
x = 6/5 - 4/5 = 2/5 or -6/5 - 4/5 = -2
Hence the three x-coordinates are
x = 2, x = - 2, and x = 2/5
The y-coordinates are derived from y = (5·x - 2)/5 as y = 8/5, y = -12/5, and y = y = 0
The coordinates of the points of intersection are (-2, -12/5), (2/5, 0), and (2, 8/5)
(b) The gradient of the curve, , is given by the differentiation of the equation of the curve, x²·y - 5·x + y + 2 = 0 which is the same as y = (5·x - 2)/(x² + 1)
Therefore, we have;
.......(3)
Which gives by plugging in the value of x in the slope equation;
At x = -2, = -0.92
At x = 2/5, = 4.3
At x = 2, = -0.28
Therefore;
Gradient at (-2, -12/5) = -0.92
Gradient at (2/5, 0) = 4.3
Gradient at (2, 8/5) = -0.28.