# Tutor profile: Ashwin A.

## Questions

### Subject: Trigonometry

Determine the exact value in radians for x, given that sin (x + pi/4) = 1/2, with 0 <=x<=2pi.

Using trigonometric identity sin (x+y) = sinxcosy + sinycosx sin (x + pi/4) = sin(x)cos(pi/4) + sin(pi/4)cos(x) pi/4 is special triangle; using ratio of 1,1, root 2, sin pi/4 = 1/root 2 and cos pi/4 = 1/root 2 sin(x)* (1/root 2) + cos(x)*(1/root 2) = 1/2 sin(x)/root 2 + cos(x)/root 2 = 1/2 (sin(x) + cos(x))/root 2 = 1/2 Square both sides: ((sin(x) + cos(x))/root 2)^2 = (1/2)^2 Expand using perfect square trinomial method - note: (x+y)^2 = x^2 + 2xy + y^2 Likewise, (sin(x) + cos(x))^2 = sin(x)^2 + 2sin(x)cos(x) + cos(x)^2 ((sinx)^2 + 2sinxcosx + (cosx)^2)/2 = 1/4 (sinx)^2 + 2sinxcosx + (cosx)^2 = 1/2 Apply Pythagorean identity: sinx^2 + cosx^2 = 1 1 + 2sinxcosx = 1/2 Substitute trigonometric identity 2sinxcosx = sin (2x) 1 + sin (2x) = 1/2 sin (2x) = -1/2 2x = sin -1 (-1/2) - use inverse sine function Using special triangle ratio of 1, root 3, 2: sin pi/3 = root 3/2, sin pi/6 = 1/2; Using CAST rule, negative sine occurs in Quad 3 and Quad 4; therefore, sin -1 (-1/2) = 7pi/6 (Quadrant 3), 11pi/6 (Quadrant 4) - (Assuming 0 <= x <= 2pi) 2x = -7pi/6, -11pi/6 x = -7pi/6 * 1/2 and x = -11pi/6 * 1/2 x = 7pi/12, 11pi/12 Proof: sub x = 7pi/12 into original equation: sin (7pi/12 + pi/4) = sin (7pi/12 + 3pi/12) = sin (10pi/12) = sin (5pi/6) = 1/2 sub x = 11pi/12 into original equation: sin (-11pi/12 + pi/4) = sin (11pi/12 + 3pi/12) = sin (14pi/12) = sin (7pi/6) = -1/2 Therefore, of the two x values 7pi/12 and 11pi/12, 7pi/12 is the only value that satisfies the equation sin (x + pi/4) = 1/2. Therefore, x = 7pi/12.

### Subject: Calculus

A ladder is being pulled up at a rate of 0.2 m per second till it reaches the foot of the roof to allow the construction worker to repair the roof shingles. The ladder is 5m long, leaning against the wall as it is pulled up to align with the roof. If the ladder reaches the roof when the wall is 4m high, how fast is the bottom of the ladder moving towards the wall?

This is an example of a problem involving related rates. A good place to start would be to create a visual to help conceptualize the problem. Visualize & Draw Diagram: The ladder leans against the wall and the ground, creating a right angled triangle. This gives rise to the pythagorean theorem, where the sum of the length of the wall and the ground distance between the bottom of the ladder and wall equal the square of the ladder length. In this case, the ladder acts as the "hypotenuse". Moving the ladder up essentially increased the length of the wall while decreasing the ground distance. There are two parameters in this situation; ground distance and wall height. Define Variables: Let "d" represent ground distance from ladder bottom to wall in meters Let "h" represent height of the wall or point at which it meets the ladder in meters Let "t" represent time in seconds The height of the ladder is 5m. 5 ^ 2 = 25 Therefore as per the Pythagorean Theorem: 25 = d^2 + h^2 Required: We are to find the rate of change of the distance from the ladder bottom toward the wall ("d") or derivative with respect to time given the rate of the ladder moving up against the wall. Given: 25 = d^2 + h^2 d (h)/dt = 0.2 m/s - this is the rate at which the height of the wall increases when the ladder moves up. Method: Related rates Find the derivative of each side of the equation with respect to time d (25)/dt = d (d^2 + h^2)/dt d(25)/dt = 0 - applying the constant rule, the derivative of a constant term is zero 0 = d(d^2 + h^2)/dt Applying the sum rule, differentiate each term added with respect to time 0 = d(d^2)/dt + d(h^2)/dt Applying the chain rule, differentiate d^2 with respect to d, multiply by derivative of d with respect to t (time). Repeat the same process for h^2. 0 = 2*d * (d(d)/dt) + 2*h* (d(h)/dt) Find the value of "d" in 25 = d^2 + h^2 at h = 4m. 25 = d^2 + 4^2 25 = d^2 + 16 d^2 = 25-16 d^2 = 9 d = 3m Given d = 3m, h = 4m, and dh/dt = 0.2m/s, plug into above equation to find d(d)/dt - rate of change of ladder towards wall: 0 = 2*3*d(d)/dt + 2*4*0.2 0 = 6 * d(d)/dt + 1.6 -1.6 = 6 * d(d)/dt d(d)/dt = -1.6/6 d(d)/dt = -0.26 m /s The rate of change of the distance of the bottom of the ladder towards the wall is -0.26 m/s when the ladder meets the wall at a 4m altitude. Please note that it is negative given that the distance of "d" is decreasing since the ladder bottom is moving closer to the wall. -

### Subject: Algebra

A rectangle has a perimeter of 64 centimetres. Its area is 135 square centimetres, while its length is 12 cm more than 3 times its width. Determine the length and width of the rectangle in question.

Given: A = 135 cm ^2 P = 64 cm length = 3 times width + 12 Define Variables: Let "w" represent the width of the rectangle. The length expressed in terms of the width is thus 3*w + 12 Required: Find the length and width. Solve for w, and determine 3w+12 (length as determined above). Determine Function: Rectangle Area = length * width Substitute terms & variables: 135 cm ^2 = w * (3w + 12) 135 = 3w^2 + 12w Rectangle Perimeter = 2 (length + width) Substitute terms & variables: 64 cm = 2 * (w + 3w + 12) * Expand and simplify by FOIL of brackets 64 = 2 * (4w + 12) 64 = 8w + 24 Divide each term on both sides by common factor of 8 to simplify 8 = w + 3 Given two equations: Perimeter: w + 3 = 8 Area: 3w^2 + 12w = 135 Determine 'w': Set equations equal to zero: w - 5 = 0 3w^2 + 12w - 135 = 0 Set equations equal to one another: w - 5 = 3w^2 + 12w - 135 Re-arrange the combined equation to set it equal to zero to set up for factoring a quadratic trinomial. 3w^2 + 12w - w - 135 + 5 3w^2 + 11w - 130 = 0 Use complex trinomial factoring: 3w^2 = 3w * w Two factors of 130 that add/subtract to 11w = 26, -5*3 (-15) (w - 5) (3w + 26) = 0 w-5 = 0 and 3w + 26 = 0 w = 5, -26/3 -26/3<0, therefore the only option for the "w" (width) value is w = 5 cm. If w = 5, substitute w = 5 into length = 3w + 12 3*(5) + 12 = 15 + 12 = 27 cm Therefore, given the above parameters in question, a rectangle of dimensions of 27cm x 5 cm would satisfy the above conditions.

## Contact tutor

needs and Ashwin will reply soon.