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Consider the closed curve between 0 <= theta < 2pi given by r(theta) = 6 + alpha sin theta, where alpha is some real constant strictly between 0 and 6. The area in this closed curve is 97pi/2. Calculate the value of alpha.

Student uses the definition of area [A = 1/2 integral r(theta)^2 d theta], and proceeds using standard integration techniques to give a quadratic solvable for alpha. [alpha^2 = 25] Thus, alpha = 5.
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Answered by Graham C. • Maths tutor
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The curve C has equation 2yx^2 + 2x + 4y - cos(πy) = 45. Using implicit differentiation, find dy/dx in terms of x and y

2x 2 y + 2x + 4y - cos(πy) = 45Applying implicit differentiation:4xy + 2x 2 (dy/dx) + 2 + 4(dy/dx) + πsin(πy)(dy/dx) = 0Moving all (dy/dx) terms to one side:2x 2 (dy/dx) + 4(dy/dx) + πsin(πy)(dy/dx) = -4xy -...
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Answered by Prahlad M. • Maths tutor
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Find the integral of [ 2x^4 - (4/sqrt(x) ) + 3 ], giving each term in its simplest form

We begin by rewriting it in a more workable form: 2x 4 - 4x -1/2 + 3. Indices are easier to integrate than fractions.Now, we integrate each term separately. The first term is 2x 4 . We increase the power of ...
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Answered by Prahlad M. • Maths tutor
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y = p x q^(x - 1), When x = 1, y = 10, and when x = 6, y = 0.3125. Find the value of 'y' when x = 3

When x = 1, y = 10Therefore, p x q (1 - 1) = 10 p x q 0 = 10 p x 1 = 10, p = 10When x = 6, y = 0.3125 p x q (6 - 1) = 0.3125 10 x q 5 = 0.3125 q 5 = 0.03125, q = (0.03125) 1/5 , q = 0.5Let x = 3, 10 x 0.5 (3...
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Answered by Prahlad M. • Maths tutor
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Write 16 × 8^2x as a power of 2 in terms of x

16 = 4 2 = (2 2 ) 2 = 2 4 8 = 2 3 , therefore 8 2x = 2 3(2x) = 2 6x 2 4 x 2 6x = 2 4 + 6x
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Answered by Prahlad M. • Maths tutor
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