Ridiculous Calculus Test

Z

zizour10

New member
Joined
May 7, 2010
Messages
2
Hey,
I am new to the forum so I am not sure on how this works... but I am in desperate need of help for my test.

We have covered all the way up to integrals, so if you know any methods to solve the following problems that are beyond integrals aren't acceptable.

Any hints/solutions/steps help. Thank you.

1) Prove that | tan x + tan y | >/ | x + y |
---(prove that the absolute value of (tan x + tan y) is greather than or equal to absolute value of (x + y) )

2) f(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e, where 2a^2 < 5b,
show that equation f(x) = 0 can't have more than three distinct real solutions (roots)

3) A light is at the top of a 16 ft. light pole. A boy 5 ft. tall walks away from the pole at the rate of 4ft/sec.
-a) at what rate is the tip of his shadow moving when he is 18ft. from the pole?
-b) at what rate is the length of his shadow increasing?

4) a rocket is rising vertically at 880 ft/sec. there is a camera on the ground recording the event. when it is 4000 ft up, how fast is the camera to rocket distance is changing at that instance? How fast must the camera's elevation angle change at that instance to keep the rocket in sight?

5) use newton's method to prove: e^x = 3-2x

6) find dy/dx if 2y^3 + (t^3)(y) = 1 and dt/dx = 1/cos t

There are plenty more, but let's start there.
Thank you so much for anyone that can help. I appreciate it!
 
zizour10 said:
Hey,
2) f(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e, where 2a^2 < 5b,
show that equation f(x) = 0 can't have more than three distinct real solutions (roots)
Click to expand...


A sure way to NOT get help is tell everyone it's a test, and then list out 6 problems.

I will be nice and help with this one.

Find \(\displaystyle f'''(x)=0\). It is 0 if \(\displaystyle 10x^{2}+4ax+b=0\). Use the quad formula to find x and then look at the discriminant.

The discriminant does not have real solutions if it is < 0. The discriminant will reduce down to something familiar in the problem statement.
 
\(\displaystyle I'll \ do \ one \ for \ you.\)

\(\displaystyle 1) \ Prove \ |tan(x)+tan(y)| \ \ge \ |x+y| \ for \ all \ reals \ in \ the \ interval \ (-\pi/2,\pi/2).\)

\(\displaystyle Mean-Value \ Theorem:\)

\(\displaystyle Let \ f(x) \ = \ tan(x) \ and \ x \ \ne \ y. \ Then, \ by \ the \ MVT, \ \frac{tan(x)-tan(y)}{x-y} \ = \ sec^2(c).\)

\(\displaystyle \frac{|tan(x)-tan(y)|}{|x-y|} \ = \ sec^2(c) \ \ge \ 1, \ |tan(x)-tan(y)| \ \ge \ |x-y|.\)

\(\displaystyle Ergo, \ replace \ y \ by \ -y \ to \ get \ |tan(x)+tan(y)| \ \ge \ |x+y|.\)
 
4) a rocket is rising vertically at 880 ft/sec. there is a camera on the ground recording the event. when it is 4000 ft up, how fast is the camera to rocket distance is changing at that instance? How fast must the camera's elevation angle change at that instance to keep the rocket in sight?
Click to expand...

This is a straightforward related rates application using a little trig and maybe Pythagoras.

Draw a right triangle and note that it is a 3-4-5 triangle (only in thousands).

Label the opposite side y, the hypoteneuse D, and the adjacent side 3000

You need dD/dt given that dy/dt=880 ft/sec.

Implicitly differentiate \(\displaystyle D^{2}=y^{2}+3000^{2}\)

Then, solve for dD/dt and plug in the given values. You're done.
 
\(\displaystyle 6) \ Find \ \frac{dy}{dx} \ if \ 2y^3+t^3y \ = \ 1 \ and \ \frac{dt}{dx} \ = \ \frac{1}{cos(t)}.\)

\(\displaystyle \frac{d}{dt}[2y^3+t^3y-1] \ = \ 3t^2y \ = \ \frac{dy}{dt}.\)

\(\displaystyle \frac{dy}{dx} \ = \ \frac{dy}{dt}\frac{dt}{dx}, \ chain \ rule, \ = \ (3t^2y)\bigg(\frac{1}{cos(t)}\bigg) \ = \ \bigg(\frac{3t^2y}{cos(t)}\bigg).\)
 
Thanks guys!
galactus said:
zizour10 said:
Hey,
2) f(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e, where 2a^2 < 5b,
show that equation f(x) = 0 can't have more than three distinct real solutions (roots)
Click to expand...


A sure way to NOT get help is tell everyone it's a test, and then list out 6 problems.

I will be nice and help with this one.

Find \(\displaystyle f'''(x)=0\). It is 0 if \(\displaystyle 10x^{2}+4ax+b=0\). Use the quad formula to find x and then look at the discriminant.

The discriminant does not have real solutions if it is < 0. The discriminant will reduce down to something familiar in the problem statement.
Click to expand...

I am still having a hard time understanding this. What is the reason of taking the 3rd derivative? And where does the "2a^2 < 5b" come into the equation?
 
You must log in or register to reply here.
Top