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Parmeshwar G.

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Linear Algebra

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Question:

. Find all solutions to the following systems of linear equations. (a) x1 − 2x2 + 2x3 = 5 x1 − x2 = −1 −x1 + x2 + x3 = 5 (b) x1 + x2 + 3x3 = 3 −x1 + x2 + x3 = −1 2x1 + 3x2 + 8x3 = 4

Parmeshwar G.

Answer:

Answer: (a) We create the augmented matrix and row reduce: 1 −2 2 5 1 −1 0 −1 −1 1 1 5 −→ 1 0 0 1 0 1 0 2 0 0 1 4 Thus, the solutions are x1 = 1, x2 = 2, x3 = 4 (b) We create the augmented matrix and row reduce: 1 1 3 3 −1 1 1 −1 2 3 8 4 −→ 1 1 3 3 0 1 2 1 0 0 0 −3 This system is inconsistent. There are no solutions.

Calculus

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Question:

Find the derivative of the following functions: a. $ f(x) = x^8 \sin 5x $ b. $ f(x) = (x + \sin x)^{23} $ c. $ f(x) = \frac{x^2 - x}{ \tan x} $ d. $ f(x) = \frac{1 + (1/x)}{1 - (1/x)} $ e. $ f(x) = (2x+1)^7 (3x+1)^5 $

Parmeshwar G.

Answer:

a. $ f(x) = x^8 \sin 5x $ Solution: $f'(x) = 8x^7 \sin 5x + x^8 (\cos 5x) 5$ by the product rule and the chain rule. b. $f(x) = (x + \sin x)^{23} $ Solution: $f'(x) = 23(x + \sin x)^{22}(1 + \cos x)$ by the chain rule. c. $ f(x) = \frac{x^2 - x}{ \tan x} $ Solution: $f'(x) = \frac{(\tan x)(2x - 1) - (\sec ^2 x) (x^2 - x)}{ \tan ^2 x} $ by the quotient rule. d. $ f(x) = \frac{1 + (1/x)}{1 - (1/x)} $ Solution: First simplify: this gives $f(x) = \frac{x + 1}{x -1} $ Now it is easy to use the quotient rule: $f'(x) = \frac{(x - 1)(1) - (x+1) (1)}{(x-1)^2} $ by the quotient rule. This can be simplfied to $ \frac{-2}{(x-1)^2} $ e. $ f(x) = (2x+1)^7 (3x+1)^5 $ Solution: $f'(x) = 7(2x+1)^6(2) (3x+1)^5 + 5(3x+1)^4(3) (2x+1)^7$ by the product and chain rules.

Algebra

TutorMe

Question:

find the solutions of these comparatively simple trigonometric equations: 1. sin(x) = cos(x). 2. sin(2x) = cos(3x). 3. sin%282x%29 = 1%2F2. 4. sin%28x%2Bpi%2F6%29 = sqrt%283%29%2F2. 5. sin%283x%2B7pi%2F11%29 = 1%2F2. 6. sqrt%282%29%2Acos%28x%29+%2B+1 = 0. 7. tan%28x%2Bpi%2F4%29 = sqrt%283%29%2F3.

Parmeshwar G.

Answer:

Solution 1 Since cos(x) cannot be zero in this problem (due to obvious reason), we can divide both sides by cos(x). Then you will get tan(x) = 1, which immediately gives the solution x = pi%2F4 or x = 5pi%2F4. These are all solutions in the interval [0,2pi). If you want to write a general solution, then x = pi%2F4+%2B+k%2Api, where k is any integer k = 0, +/-1, +/-2, . . . Solution 2 Use the formula sin%28x%29 = cos%28pi%2F2-x%29. Then the equation takes the form cos%28pi%2F2-x%29 = cos%28x%29. If you consider the solutions "x" in the interval [0,2pi), then it implies pi%2F2+-+x = x ---> pi/2 = 2x ---> x = pi%2F4, or 2pi+%2B+pi%2F2+-+x = x ---> 5pi%2F2 = 2x ---> x = 5pi%2F4. Again, if you want to write a general solution, then x = pi%2F4+%2B+k%2Api, where k is any integer k = 0, +/-1, +/-2, . . . Problem 2 Solve an equation sin(2x) = cos(3x). Solution The method of the Solution 1 above to the previous problem doesn't work in this case. But you still can start using the formula sin%282x%29 = cos%28pi%2F2-2x%29. Then the equation takes the form cos%28pi%2F2-2x%29 = cos%283x%29. If you consider the solutions "x" in the interval [0,2pi), then it implies pi%2F2+-+2x = 3x ---> pi/2 = 5x ---> x = pi%2F10, (1) or 2pi+%2B+pi%2F2+-+2x = 3x ---> 5pi%2F2 = 3x+%2B+2x ---> 5x = 5pi%2F2 ---> x = 5pi%2F10, (2) or 4pi+%2B+pi%2F2+-+2x = 3x ---> 9pi%2F2 = 3x+%2B+2x ---> 5x = 9pi%2F2 ---> x = 9pi%2F10, (3) or 6pi+%2B+pi%2F2+-+2x = 3x ---> 13pi%2F2 = 3x+%2B+2x ---> 5x = 13pi%2F2 ---> x = 13pi%2F10, (4) or 8pi+%2B+pi%2F2+-+2x = 3x ---> 17pi%2F2 = 3x+%2B+2x ---> 5x = 17pi%2F2 ---> x = 17pi%2F10. (5) By adding another 2pi to the very left side of (5), you will see that the values of x start repeating cyclically with the period 2pi and do not produce new solutions. If you want to write a general solution, you can do it in the form x = pi%2F10+%2B+k%2A%282pi%2F5%29, where k is any integer k = 0, +/-1, +/-2, . . . Problem 3 Solve an equation sin%282x%29 = 1%2F2. Solution Since the sine is equal to 1%2F2, the argument of sine is pi %2F6 or 5pi%2F6. In other words, either 2x = pi%2F6, which implies x = pi%2F12, or 2x = 5pi%2F6, which implies x = 5pi%2F12. If you want to write a general solution, then either 2x = pi%2F6+%2B+2k%2Api, which implies x = pi%2F12+%2B+k%2Api, or 2x = 5pi%2F6+%2B+2k%2Api, which implies x = 5pi%2F12+%2B+k%2Api, where k is any integer k = 0, +/-1, +/-2, . . . Problem 4 Solve an equation sin%28x%2Bpi%2F6%29 = sqrt%283%29%2F2. Solution Since the sine is equal to sqrt%283%29%2F2, the argument of sine is pi %2F3 or 2pi%2F3. In other words, either x+%2B+pi%2F6 = pi%2F3, which implies x = pi%2F3+-+pi%2F6 = pi%2F6, or x+%2B+pi%2F6 = 2pi%2F3, which implies x = 2pi%2F3+-+pi%2F6 = 3pi%2F6 = pi%2F2. If you want to write a general solution, then x = pi%2F6+%2B+2k%2Api and/or x = pi%2F2+%2B+2k%2Api, where k is any integer k = 0, +/-1, +/-2, . . . Problem 5 Solve an equation sin%283x%2B7pi%2F11%29 = 1%2F2. Solution Since the sine is equal to 1%2F2, the argument of sine is pi %2F6 or 5pi%2F6. In other words, either 3x+%2B+7pi%2F11 = pi%2F6, which implies x = pi%2F6+-+7pi%2F11 = 11pi%2F66+-+42pi%2F66 = -31pi%2F66, or 3x+%2B+7pi%2F11 = 5pi%2F6, which implies x = 5pi%2F6+-+7pi%2F11 = 55pi%2F66+-+42pi%2F66 = 13pi%2F66. If you want to write a general solution, then x = -31pi%2F66+%2B+%282k%2Api%29%2F3 and/or x = 13pi%2F66+%2B+%282k%2Api%29%2F3, where k is any integer k = 0, +/-1, +/-2, . . . Problem 6 Solve an equation sqrt%282%29%2Acos%28x%29+%2B+1 = 0, for theta <= x < 2pi. Solution The equation sqrt%282%29%2Acos%28x%29+%2B+1 = 0 is equivalent to cos(x) = -1%2Fsqrt%282%29, or cos%28x%29 = -sqrt%282%29%2F2. Hence, x = 3pi%2F4 and/or x = 5pi%2F4. These are only solutions in the given interval. Answer. The solutions are x = 3pi%2F4 and/or x = 5pi%2F4. Problem 7 Solve an equation tan%28x%2Bpi%2F4%29 = sqrt%283%29%2F3. Solution Since the tangent is equal to sqrt%283%29%2F3, the argument of the tangent is pi %2F3 or 4pi%2F3. In other words, either x+%2B+pi%2F4 = pi%2F3, which implies x = pi%2F3+-+pi%2F4 = pi%2F12, or x+%2B+pi%2F4 = 4pi%2F3, which implies x = 4pi%2F3+-+pi%2F4 = 13pi%2F12. If you want to write a general solution, then x = pi%2F12+%2B+2k%2Api with the period pi, where k is any integer k = 0, +/-1, +/-2, . . .

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