# Tutor profile: Landon W.

## Questions

### Subject: Calculus

Find the domain and range of f(x, y) = sqrt( 9 - x^2 - y^2).

Recall that we cannot take the square root of a negative number and the equation of a circle ----> (x - h)^2 + (y - k)^2 = r^2. Since (9 - x^2 - y^2) is underneath a square root we would set this >= 0. After some algebraic manipulation we get x^2 + y^2 <= 9, which is the domain of f(x,y). Now by looking at this equation we can see that it is also the equation of a circle with center at (0,0) and a radius of 3. Therefore our range is from [0, 3].

### Subject: Calculus

Evaluate the integral S cos^3(x) dx.

The first thing you would think to do is a u substitution for u = cos(x), but this isn't very helpful since du = -sin(x) dx. Recall the trig identify sin^2(x) + cos^2(x) = 1. Since we have cos^3(x) we can break this up into an equivalent form of cos^2(x)*cos(x). Once we have broken it up we can solve our trig identity using simple algebra for cos^2(x) which would be cos^2(x) = 1 - sin^2(x). We can now plug this into our integral for cos^2(x) which makes our integral look like this ----> S (1 - sin^2(x)) * cos(x)). We can now evaluate using u substitution by letting u = sin(x) with du = cos(x)dx. Now putting the integral in terms of u we would write ----> S (1 - u^2)du . Now we have a simple integral that can be solved using the power rule which equals (u - (u^3/3) + C). All we have left to do now is plug in our value we set equal to u and our final answer is (sin(x) - (sin^3(x) / 3) + C).

### Subject: Calculus

Determine the intervals on which the function (f(x) = (x^2 - 4) / (x^2 - 8x - 20)) is continuous. If the function has discontinuities, determine whether if removable or non-removable.

At first glance we can see that the numerator (x^2 - 4) can be factored into (x - 2)(x + 2), and that the denominator (x^2 - 8x - 20) can be factored into (x - 10)(x + 2). Since the problem asks for continuous intervals of the function we must first check for the domain of the function. To find the domain of a fraction, we must find where the denominator equals zero. Solving the factored denominator we see the function is discontinuous at x = -2, and x = 10. In interval notation we can write that our interval of continuity is (-infinity, -2) U (-2, 10) U (10, +infinity). Since there is a discontinuity we must determine whether it is removable or non-removable. By looking at our factored function we see that the function has a matching factor in the numerator and denominator --> (x + 2). This means we have a removable discontinuity or hole at x = -2 because the function is undefined on the line at that certain point. Our other discontinuity was at x = 10. Since this point does not lie on the line and in the factored numerator it's considered a non-removable discontinuity.

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