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Dan W.

Former University Instructor and tutor of over 10 years

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Pre-Calculus

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Question:

Prove that $$sin^2\theta + cos^2\theta = 1$$

Dan W.

Answer:

Imagine a circle in the x-y plane centered at the origin. You can draw a line from the origin to the edge of this circle. The length of that line will be the radius of the circle, which we can call $$r$$, as it can have any arbitrary value. Drawing this line from the origin to the edge of the circle will also create an angle with the horizontal x-axis, we may call this angle $$\theta$$. The point to which we've drawn our line has a pair of (x,y) coordinates. If we then draw a line straight down to the x-axis, and the straight back to the origin, we will have created a right triangle. This right triangle will have a base length of x, and a leg height of y (the same x and y from our earlier (x,y) pair), and a hypotenuse of $$r$$, the radius of our circle. By the Pythagorean theorem $$x^2 + y^2 = r^2$$ that is, the sum of the square of the two sides of a right triangle is equal to the square of its hypotenuse. Additionally, since this is a right triangle with opposite side y and adjacent side x, we may express x and y in the following way $$cos\theta = \frac{x}{r}$$ $$sin\theta = \frac{y}{r}$$ or $$x = rcos\theta$$ $$y = rsin\theta$$ so by pythagorean theorem $$x^2 + y^2 = r^2$$ becomes $$(rcos\theta)^2 + (rsin\theta)^2 = r^2$$ $$r^2cos^2\theta + r^2sin^2\theta = r^2$$ factoring out an $$r^2$$ on the left gives $$r^2(cos^2\theta + sin^2\theta) = r^2$$ dividing both sides by $$r^2$$ gives $$cos^2\theta + sin^2\theta = 1$$ or rather $$sin^2\theta + cos^2\theta = 1$$

Calculus

TutorMe

Question:

Find the equation of the tangent line for the function $$f(x) = e^{sinx}$$ at the point x = 0

Dan W.

Answer:

First, it is best to evaluate the function at the point x = 0. $$f(0) = e^{sin0} = e^{0} = 1$$ So we are talking about the tangent line at the point (0,1) To find the equation of the tangent line, you need to first find the first derivative of the function $$f(x) = e^{sin2x}$$. This will require chain rule. Chain rule states that if we can find a function $$u(x)$$ such that $$f(x) = f(u(x))$$ then the first derivative of $$f(x)$$ can be expressed as $$\frac{df}{dx} = \frac{df}{du}\frac{du}{dx}$$ So we just need to make an appropriate choice of $$u(x)$$. The best choice here is to be $$u = sinx$$ which gives us $$ f(u) = e^{u}$$ this gives $$\frac{du}{dx} = cosx$$ and $$\frac{df}{du} = e^{u}$$ so the derivative, by chain rule is $$\frac{df}{dx} = \frac{df}{du}\frac{du}{dx} = cosxe^{u} = cosxe^{sinx}$$ Now that we have the first derivative of the function, we can calculate the slope of the tangent line at x = 0. Remember that the slope of the tangent line is just the first derivative evaluated at the given x value. So we just need to plug x = 0 into our first derivative giving $$m_{tan} = cos(0)e^{sin0} = 1*e^0 = 1*1 = 1$$ So we now have all the pieces. The equation for a tangent line is of course $$y - y_0 = m_{tan}(x - x_0)$$ with $$x_0$$ and $$y_0$$ being our initial x and $$f(x)$$ values for x = 0, making those substitutions gives $$y - 1 = 1(x - 0)$$ or $$y = x + 1$$

Algebra

TutorMe

Question:

Solve the following system of equations for x and y by substitution x - 4y = -18 x - 3y = -11

Dan W.

Answer:

This can be solved either via substitution or by elimination. Substitution: first, solve the top equation for x in terms of y. What I mean by this is find a way to get x by itself into the equation $$ x - 4y = -18$$ First, you will want to eliminate the $$-4y$$ from the left-hand side of the equation. Since it is in the form of a subtraction, you will need to add $$4y$$ to both sides $$x - 4y = -18$$ $$+4y$$ $$+4y$$ which then gives $$x = -18 + 4y$$ or $$x = 4y - 18$$ So now we have x solved in terms of y. Of course, we could have just as easily reworked the equaiton $$x - 4y = -18$$ to get y by itself, but in this case it was easier to get x. So just as a reminder, we started with the two equations 1) $$x - 4y = -18$$ 2) $$x - 3y = -11$$ Using our substitution of $$x = 4y - 18$$ which we got from the top equation, we can make a substitution for x into the second equation which gives $$(4y-18) - 3y = -11$$ Notice that we simply substitured $$x$$ with $$4y-18$$ because we know that $$x = 4y-18$$. This gives us $$4y - 18 - 3y = -11$$ combining the y terms gives $$4y - 3y - 18 = -11$$ $$1y - 18 = -11$$ $$y - 18 = -11$$ and we solve for y by simply adding 18 to both sides $$y - 18 = -11$$ $$+18$$ $$+18$$ $$y = 7$$ substituting this y value in the first equation gives $$x - 4(7) = -18$$ $$x - 28 = -18$$ this equation can be solved for x by adding 28 to both sides $$x - 28 = -18$$ $$+28$$ $$+28$$ giving $$x = 10$$ So we have the solutions x = 10 and y = 7. As a sanity check, plug these values into either of the two given equations to see if they work.

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