Analytic Geometry And Calculus Mat 102 Test
Questions
1(a) Evaluate the
integral
(1/ √x(1 + √x)²) × dx
(b) Determine the equation of the tangent and the normal to the circle
x² + y² - 2y + 6x-7= 0 at the point P(-2,5)
(c) Given 2x³-3y² = 7,
show that y³y" - 2xy² + x⁴=0.
(Note: find the second derivative)
2(a) Evaluate the integral
(In(x + 2))/(2x +4)) ×dx
(b) The straight line y = x+4 cuts the circle x² +y² = 26.at A and B
I. Calculate the coordinates of A and B.
II. Determine the coordinates of M, the mid-point of chord AB.
III. Determine the equations of the tangents to the circle at A and B
IV. Determine the coordinates of S, the point where the two tangent intersect.
(c) If x³+y³ = 7,
show that y⁵y" +14x = 0
(Note find the second derivative)
SOLUTIONS
Provide explanations for each answer to the given questions.
1(a) Evaluate the integral (1/ √x(1 + √x)²) × dx.
Let's use the substitution method, where u = √x, so x = u². Then du = dx/(2√x) = dx/(2u).
Now, rewrite the integral in terms of u:
∫ (1/ u(1 + u)²) × (2u)du = 2∫ (1/ (1 + u)²) du.
Now we can solve the integral:
2∫ (1/ (1 + u)²) du = 2 (-1/(1+u)) + C = -2/(1 + √x) + C.
Answer: -2/(1 + √x) + C.
(b) Determine the equation of the tangent and the normal to the circle x² + y² - 2y + 6x-7= 0 at the point P(-2,5).
Let's rewrite the equation of the circle in standard form by completing the square:
(x² + 6x) + (y² - 2y) = 7
(x² + 6x + 9) + (y² - 2y + 1) = 17
(x + 3)² + (y - 1)² = 17
The circle has the center at (-3, 1) and radius of sqrt(17).
To find the slope of the tangent at point P(-2, 5), we find the slope of the line connecting the center and point P:m_CP = (5 - 1) / (-2 - (-3)) = 4.
The slope of the tangent is the negative reciprocal of m_CP:
m_tangent = -1 / 4.
Now we use the point-slope form for the tangent line:
y - 5 = -1/4 (x + 2) => y = -1/4x - 1/2 + 5 => y = -1/4x + 9/2.
The slope of the normal line is the negative reciprocal of the tangent line's slope:
m_normal = 4.
Using the point-slope form for the normal line:
y - 5 = 4(x + 2) => y = 4x + 13.
Equation of the tangent: y = -1/4x + 9/2
Equation of the normal: y = 4x + 13
(c) Given 2x³-3y² = 7, show that y³y" - 2xy² + x⁴=0. (Note: find the second derivative)
To find the second derivative, let's start with the first derivative:
2x³ - 3y² = 7
Differentiate with respect to x:
6x² - 6yy' = 0.
Now, isolate y':
y' = x²/y.
Now, differentiate y' with respect to x to find y":
y" = d/dx(x²/y)
Using the Quotient rule for derivatives, (u/v)' = (vu' - uv')/v², where u = x² and v = y:
y" = (y(2x) - x²(2yy'))/y²
y" = (2xy - 2x³y')/y²
Substitute y' = x²/y from our previous result:
y" = (2xy - 2x³(x²/y))/y²
y" = (2xy - 2x⁵/y)/y²
Now, substitute y' and y" into the required expression to prove it equal to zero:
y³y" - 2xy² + x⁴ = 0
y³(2xy - 2x⁵/y)/y² - 2xy² + x⁴ = 0
(2xy⁴ - 2x⁵y)/y² - 2xy² + x⁴ = 0
Multiply the equation by y² to eliminate the denominator:
2xy⁴ - 2x⁵y - 2xy⁴ + x⁴y² = 0
Combine like terms:
- 2x⁵y + x⁴y² = 0
Factor out a common factor x:
x(- 2x⁴y + x³y²) = 0
Since x cannot be 0, the equation holds true, and we have proven that
y³y" - 2xy² + x⁴ = 0.
2(a) Evaluate the integral:
∫(ln(x + 2))/(2x + 4) dx
First, notice that by making a substitution, the integral becomes easier:
Let u = x + 2, then du = dx
The integral becomes:
∫(ln(u))/(2(u-2) + 4) du
Now, simplify the integral:
∫(ln(u))/(2u) du
We can use integration by parts to solve this:
Let v = ln(u), then dv = du/u
Let w = 1/2u, then dw = -1/2u² du
Using integration by parts, ∫v * dw = vw - ∫w * dv:
∫(ln(u))/(2u) du = (1/2)u ln(u) - ∫(1/2u * du/u)
= (1/2)u ln(u) - ∫(1/2u) du = (1/2)u ln(u) - (1/2)∫(du)
Now, integrate and substitute back x + 2 for u:
(1/2)(x + 2) ln(x + 2) - (1/2)x + C
So the integral evaluates to:
(1/2)(x + 2) ln(x + 2) - (1/2)x + C
I. To find the coordinates of A and B, we substitute y = x+4 into the equation of the circle:
x² + (x+4)² = 26
Simplifying this gives:
2x² + 8x - 10 = 0
Solving for x using the quadratic formula gives:
x = (-8 ± √104)/4
x ≈ -2.73 or x ≈ 1.23
Substituting these values into y = x+4 gives the coordinates of A and B:
A ≈ (-2.73, 1.27)
B ≈ (1.23, 5.27)
II. The midpoint M of AB is the average of A and B:
M = ((-2.73 + 1.23)/2, (1.27 + 5.27)/2)
M ≈ (-0.75, 3.27)
III. The slope of the line y = x+4 is 1, so the slopes of the tangent lines to the circle at A and B are the negative reciprocals of 1, which is -1. We use the point-slope form of the equation for a line, y-y1 = m(x-x1), where (x1, y1) is the point of tangency and m is the slope of the tangent line.
At A, the slope of the tangent line is -1, and the point of tangency is A ≈ (-2.73, 1.27), so we have:
y - 1.27 = -1(x + 2.73)
y = -x - 1.46
At B, the slope of the tangent line is -1, and the point of tangency is B ≈ (1.23, 5.27), so we have:
y - 5.27 = -1(x - 1.23)
y = -x + 6.5
IV. To find the coordinates of the intersection point S, we set the two tangent line equations equal to each other and solve for x:
-x - 1.46 = -x + 6.5
7.96 = 2x
x ≈ 3.98
Substituting this value of x into one of the tangent line equations gives the y-coordinate:
y ≈ 2.52
Therefore, the coordinates of S are S ≈ (3.98, 2.52)
C. Given x³+y³ = 7
Taking the first derivative with respect to x, we get:
3x² + 3y² dy/dx = 0
dy/dx = -x²/y² …(1)
Taking the second derivative with respect to x, we get:
d²y/dx² = d/dx(-x²/y²)
Using the quotient rule,
= [-2y(-2xy) - (-x²)(-2y²dy/dx)]/y^4
= (4xy/y^4) - (2x²/y³)(-x²/y²)
= 4x/y^3 + 2x^4/y^5
= 2x (2/y³ + x³/y⁵) …(2)
Multiplying equation (1) by x and substituting it in equation (2),
d²y/dx² = 2x(-xy²/x² + x³/y⁵)
= -2xy + 2x⁴/y⁵
= -2x(y - x²/y⁴)
Substituting x³+y³ = 7, we get y³ = 7-x³
Plugging this in the above equation, we get:
d²y/dx² = -2x(y - x²/(7-x³)⁴)
= -2x(y(7-x³)⁴ - x²)/(7-x³)⁴
= -2x(y(7-x³)⁴ - x²(7-x³))/(7-x³)⁴
= -2x(7y - yx³ - 7x² + x⁴)/(7-x³)⁴
= (-14xy + yx⁴ + 14x²/(7-x³)⁴)
Multiplying both sides by y⁵,
y⁵d²y/dx² = (-14x)y⁴ + (yx⁴ + 14x²)y
y⁵d²y/dx² + 14x = -yx⁴
Therefore, y⁵y" +14x = 0.
Hence, the given statement is proved.
