# Calculus Test Fixes

• Jan 13th 2010, 04:25 PM
DarkestEvil
Calculus Test Fixes
6.b. Consider the curve defined by the equation: $\displaystyle y+\cos y=x+1$ for $\displaystyle 0<y<2\pi$
Write an equation for each vertical tangent to the curve. (Hint: Vertical lines are x=#)

-I already found that $\displaystyle y'=\frac{1}{1-\sin y}$, but im not sure how to solve this one. how am i supposed to start off

10. The expression $\displaystyle \frac{1}{4}(\sqrt{1}+2\sqrt{\frac{5}{4}}+2\sqrt{2} +2\sqrt{\frac{13}{4}}+\sqrt{5})$ is the trapezoid approximation for which of the following definite integrals?

This kind of expression always baffles me, it would really help if someone would care to explain

2.a. Given that f is the function defined by $\displaystyle f(x)=\frac{x^3-x}{x^3-4x}$
Find limit goes to infinity of f(x)

-limits have always confused me. what am i supposed to be finding?

2.b. Write an equation for each vertical and each horizontal asymptote to the graph of f.

-i found that the vertical asymptotes are at x=0, x=2, x=-2 and horizontal at y=1, at least thats what i think. what am i supposed to with these?

• Jan 13th 2010, 04:39 PM
skeeter
Quote:

Originally Posted by DarkestEvil
6.b. Consider the curve defined by the equation: $\displaystyle y+\cos y=x+1$ for $\displaystyle 0<y<2\pi$
Write an equation for each vertical tangent to the curve. (Hint: Vertical lines are x=#)

-I already found that $\displaystyle y'=\frac{1}{1-\sin y}$, but im not sure how to solve this one. how am i supposed to start off

vertical tangent lines occur where the derivative is undefined ...
1 - sin y = 0 ... solve for y

10. The expression $\displaystyle \frac{1}{4}(\sqrt{1}+2\sqrt{\frac{5}{4}}+2\sqrt{2} +2\sqrt{\frac{13}{4}}+\sqrt{5})$ is the trapezoid approximation for which of the following definite integrals?

This kind of expression always baffles me, it would really help if someone would care to explain

it might help to see the "integrals"

2.a. Given that f is the function defined by $\displaystyle f(x)=\frac{x^3-x}{x^3-4x}$
Find limit goes to infinity of f(x)

-limits have always confused me. what am i supposed to be finding?

the horizontal asymptote ... divide every term by the highest power of x and then take the limit as x goes to infinity

2.b. Write an equation for each vertical and each horizontal asymptote to the graph of f.

-i found that the vertical asymptotes are at x=0, x=2, x=-2 and horizontal at y=1, at least thats what i think. what am i supposed to with these?

this part is correct except x = 0 is not a vertical asymptote ... it is a removable discontinuity (a hole)

...
• Jan 13th 2010, 04:54 PM
DarkestEvil
6.b. Oh I see, so in order for it to be undefined, y must equal $\displaystyle \frac{\pi}{2}$!

10. I'm sorry the answer choices are:
a. $\displaystyle \int_{1}^{5}\sqrt{x} dx$
b. $\displaystyle \int_{0}^{4}\sqrt{x^2+1} dx$
c. $\displaystyle \int_{0}^{2}\sqrt{x^2+1} dx$
d. $\displaystyle \int_{-1}^{2}\sqrt{x^2+1} dx$
e. $\displaystyle \int_{1}^{3}\sqrt{x} dx$

2.a. Oh so since the x with the highest power on the top and bottom are $\displaystyle x^3, \frac{x^3}{x^3}=1$!

2.b. I see your logic, so how am I supposed to write an equation for each of these values?
• Jan 13th 2010, 04:57 PM
Prove It
Quote:

Originally Posted by DarkestEvil
6.b. Oh I see, so in order for it to be undefined, y must equal $\displaystyle \frac{\pi}{2}$!

10. I'm sorry the answer choices are:
a. $\displaystyle \int_{1}^{5}\sqrt{x} dx$
b. $\displaystyle \int_{0}^{4}\sqrt{x^2+1} dx$
c. $\displaystyle \int_{0}^{2}\sqrt{x^2+1} dx$
d. $\displaystyle \int_{-1}^{2}\sqrt{x^2+1} dx$
e. $\displaystyle \int_{1}^{3}\sqrt{x} dx$

2.a. Oh so since the x with the highest power on the top and bottom are $\displaystyle x^3, \frac{x^3}{x^3}=1$!

2.b. I see your logic, so how am I supposed to write an equation for each of these values?

Actually, $\displaystyle y = \frac{\pi}{2} + 2\pi n$ where $\displaystyle n$ is an integer representing how many times you have gone around the unit circle.
• Jan 13th 2010, 05:04 PM
DarkestEvil
Quote:

Originally Posted by Prove It
Actually, $\displaystyle y = \frac{\pi}{2} + 2\pi n$ where $\displaystyle n$ is an integer representing how many times you have gone around the unit circle.

well played, thanks
• Jan 13th 2010, 05:43 PM
skeeter
Quote:

Originally Posted by Prove It
Actually, $\displaystyle y = \frac{\pi}{2} + 2\pi n$ where $\displaystyle n$ is an integer representing how many times you have gone around the unit circle.

the problem statement says $\displaystyle 0 < y < 2\pi$
• Jan 13th 2010, 05:44 PM
Prove It
Quote:

Originally Posted by skeeter
the problem statement says $\displaystyle 0 < y < 2\pi$

Ah, true.

I was just giving our OP a valuable lesson in "reading the question properly" (Giggle)