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Kathleen K.
Math Teacher & Tutor for 25 years and I love it still!!
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Pre-Calculus
TutorMe
Question:

Line L is tangent to the circle x² + y² - 2x + 8y = 8 at the point (4, y). Line L does not lie in the third quadrant. Please find the equation of Line L.

Kathleen K.
Answer:

First, let's complete the square to get the circle into its standard form: x² - 2x + ___ + y² + 8y + ___ = 8 x² - 2x + (-1)² + y² + 8y + (4)² = 8 + (-1)² + (4)² (x - 1)² + (y + 4)² = 25 The center of this circle is (1, -4) with radius 5. The point of tangency is a bit of a mystery, given to be (4, y). To solve for the missing y, let us plug in x = 4: (4 - 1)² + (y + 4)² = 25 9 + (y + 4)² = 25 (y + 4)² = 16 Square root both sides of the equation: y + 4 = $$\pm$$4 y = -4 + $$\pm$$4 y = -4 + 4 = 0 or y = -4 - 4 = -8 So there are two points of tangency: (4,0) or (4, -8). Looking at a rough sketch of the circle, we see that the tangent line at (4,0) will have a negative slope and never intrude into QIII, while the tangent line at (4, -8) will have a positive slope and definitely pass into QIII. Therefore, the proper tangent line is the one containing (4,0). What is the slope of this line? Recall from geometry that a radius drawn to a point of tangency is perpendicular to that tangent line. So if we determine the slope of the radius from center (1, -4) to (4, 0), we can then use the perpendicular slope to the radius. From the slope formula, the radius has slope $$\frac{-4 - 0}{1-4}$$ = 4/3. Therefore the tangent line has a slope of -3/4, the opposite reciprocal. Finally, using the point-slope equation of a line with the point (4, 0) and m = -3/4: y - 0 = $$-\frac{3}{4}$$(x - 4) y = $$-\frac{3}{4}x + 3$$

Trigonometry
TutorMe
Question:

The 17-inch diameter tires on a Nissan Sentra make 1000 revolutions per minute. (a) What is the angular velocity in radians per second. (b) How fast, in miles per hour, is the Sentra traveling at that time?

Kathleen K.
Answer:

I find that dimensional analysis (also known as "unit conversions") works best for angular and linear velocity questions in trigonometry. We can use the following string of equalities in any pair: 1 revolution = 2$$\pi $$ radians = 360° = 1 circumference (2$$\pi $$r units) The first three in the above string fall are categorized as "angular velocity." The radius of the rotation has no bearing on these angular velocities. As soon as you bring in circumference (with its units of length like inches, centimeters, miles, etc.), then you are in the category of "linear velocity." (a) $$\frac{1000\,revolutions}{1\, minute} \cdot \frac{2\pi\,radians}{1\,revolution}\cdot \frac{1\,minute}{60\,seconds} = \frac{2000\pi\,radians}{{60\,seconds}} = \frac{\frac{100\pi}{3}\,radians}{1\,second} $$ (b) The circumference of the 17-inch diameter tire is 17$$\pi$$ inches. Therefore, we can use the unit converter $$\frac{1\,circumference}{1\,revolution} \rightarrow \frac{17\pi\,inches}{1\,revolution} $$. We will also use the conversion 5280 feet = 1 mile. $$\frac{1000\,revolutions}{1\, minute} \cdot \frac{17\pi\,inches}{1\,revolution}\cdot\frac{1\, foot}{12\, inches}\cdot\frac{1\, mile}{5280\, feet}\cdot\frac{60\, minutes}{1\, hour} \approx$$ 50.6 miles per hour Notice in part (b), there are a total of four units canceling out from a numerator and a denominator: revolutions, inches, feet, and minutes, leaving us with miles in the numerator and hours in the denominator, which is the requested speed in miles per hour!

Calculus
TutorMe
Question:

Find the average value of the function f(x) = x$$ \sqrt{x-2} $$ on the interval [6,11].

Kathleen K.
Answer:

The average value of a function f(x) on [a,b] is defined as: $$ \frac{1}{b-a} \int_a^b f(x) dx $$ Here we have $$ \frac{1}{11-6} \int_6^{11} x\sqrt{x - 2}\,\, dx = \frac{1}{5} \int_6^{11} x\sqrt{x-2}\,\, dx $$ This beckons us with a nice u-substitution, as follows: Let u = x - 2 $$\frac{du}{dx} = 1 \rightarrow du = dx $$ If u = x - 2, then u + 2 = x Recall that the integral's lower boundaries are also in terms of x, specifically x = 6 and 11. Using u = x - 2 when x = 6 yields u = 4. Similarly, when x = 11, u = 9. We now have everything we need to transform the definite integral that is currently in terms of x into one that uses the variable u: Given: $$ \frac{1}{5} \int_6^{11} x\sqrt{x-2}\,\, dx $$ Make all the u-substitutions: = $$ \frac{1}{5} \int_4^{9} (u + 2)\sqrt{u}\,\, du $$ Rewrite the square root in exponential form: = $$ \frac{1}{5} \int_4^{9} (u + 2)u^{1/2}\,\, du $$ Distribute: = $$ \frac{1}{5} \int_4^{9} (u^{3/2} + 2u^{1/2})\,\, du $$ Integrate using the reverse power rule: = $$ \frac{1}{5} [\frac{2}{5}u^{5/2} + \frac{4}{3}u^{3/2}]$$, from u = 4 to 9 Plug in u = 9 minus the result when u = 4: = $$ \frac{1}{5} [\frac{2}{5}9^{5/2} + \frac{4}{3}9^{3/2} - \frac{2}{5}4^{5/2} - \frac{4}{3}4^{3/2}]$$ = $$\frac{1}{5} [\frac{2}{5}\cdot 3^{5} + \frac{4}{3}\cdot 3^{3} - \frac{2}{5}\cdot 2^{5} - \frac{4}{3}\cdot 2^{3}]$$ = $$\frac{1646}{75}$$

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