Solving the Quadratic Equation y^2 = 4(x+4)^3 for x in [0,2]

[suspenc3]

$$y^2 = 4(x+4)^3 0<=x<=2 ...y>0$$

Ok..so do I just simplify it to y=etc... and then find the derivative and solve..I tried this and I get a kinda long derivative..just seems like it should be easier?

Thanks

 

[HallsofIvy]

$$y^2 = 4(x+4)^3 0<=x<=2 ...y>0$$

Ok..so do I just simplify it to y=etc... and then find the derivative and solve..I tried this and I get a kinda long derivative..just seems like it should be easier?

Thanks

Not all that "long" is it? the derivative of y², with respect to x, is $\2y \frac{dy}{dt}$ and the derivative of 4(x+4)³ is 12(x+ 4)² so differentiating both sides you get
$$2y \frac{dy}{dx}= 12(x+4)^2$$
So what is $\frac{dy}{dx}$?

 

[suspenc3]

oh, i did it completely wrong..i took the root of both sides and hen tried to get the derivative

 

[benorin]

Whether you solve it implicitly (HallsofIvy's post) or explicitly (taking the rott, etc.) the result is the same: the implicit method gives

$$2y \frac{dy}{dx}= 12(x+4)^2\Rightarrow \frac{dy}{dx}= \frac{6(x+4)^2}{y} = \frac{6(x+4)^2}{2(x+4)^{\frac{3}{2}}}=3(x+4)^{\frac{1}{2}}$$

since solving the given equation for y gives $$y=2(x+4)^{\frac{3}{2}$$ and the positive root was chosen since y>0. This is the starting point for the explicit method, just differentiate the last expression

$$y=\frac{d}{dx}\left[ 2(x+4)^{\frac{3}{2}}\right] = 3(x+4)^{\frac{1}{2}}$$

which is actually easier.

 

[suspenc3]

oohhh..Ok Thanks, I get it

 

[benorin]

start with u=x+4, then try a trig substitution

 

[suspenc3]

cant i just do..$$\int_0^2 \sqrt{1+(3(x+4)^1^/^2)^2}$$

$$\int_0^2 \sqrt{37+9x}$$

 

[benorin]

cant i just do..$$\int_0^2 \sqrt{1+(3(x+4)^1^/^2)^2}$$

$$\int_0^2 \sqrt{37+9x}$$

Yes, absolutely: I was in error.

 

[suspenc3]

but wha tyou have there..is $$\sqrt{3(x+4)}^2 = 9(x+4)^2$$..

shouldnt it be: $$\sqrt{3(x+4)}^2 = 8(x+4)$$..which can be simplified and then its an easy substitution.

 

[suspenc3]

oh,,ok..Thanks!

 

[benorin]

use rather u=37+9x so that du=9dx to get

$$\int_0^2 \sqrt{37+9x} dx = \frac{1}{9}\int_{37}^{55} \sqrt{u}du$$

 

[suspenc3]

how about this one:

$$y= \frac{x^5}{5} + \frac{1}{10x^3} 1>=x>=2$$

I found the derivative to be $$\frac{5x^4}{6} - \frac{3}{10x^4}$$..im kinda stuck here..should I just keep going and will I eventually see a substituion?

 

[suspenc3]

yes that is what i got..except du=9dx

 

[benorin]

yep, it was

 

[suspenc3]

if I have $$\int_0^\pi^/^3 sqrt{sec^2x}$$..should I substitute u = sec^2x..or can it just be secx..and find the integral?

 

[pizzasky]

Do you mean $$\int_0^\frac{\pi}{3}\sqrt{sec^2x} dx$$?

If yes, note that since secx is positive from 0 to $$\pi$$/3, then $$\sqrt{sec^2x}$$ is equivalent to secx.

There is a standard formula for evaulating $$\int_0^\frac{\pi}{3}secxdx$$. Do you know what it is?

 

[benorin]

how about this one:

$$y= \frac{x^5}{5} + \frac{1}{10x^3} 1>=x>=2$$

I found the derivative to be $$\frac{5x^4}{6} - \frac{3}{10x^4}$$..im kinda stuck here..should I just keep going and will I eventually see a substituion?

The derivative of $$y= \frac{x^5}{5} + \frac{1}{10x^3}$$ is $$y^{\prime}= x^4 - \frac{3}{10x^4}$$

 

[suspenc3]

ohh..sorry its suppose to be $$y= \frac{x^5}{6} + \frac{1}{10x^3} 1>=x>=2$$

 

[suspenc3]

and for pizzasky..no I don't know how..I found some stuff on the internet concerning the integral of secx..I never knew this integral up until now..so I am guessing there's another way

 

[pizzasky]

Visit this website to see how you can integrate secx with respect to x.

http://www.math.ubc.ca/~feldman/m121/secx.pdf

All the best!

 

[suspenc3]

Oh..That explains it..so when I get the integral it is going to be

$$ln|sec \pi /3tan \pi /3| -ln|sec0tan0$$..the second part isn't going to exist will it..how is this going to afffect my answer?

 

[pizzasky]

Read carefully... Are you supposed to multiply the secant and tangent terms for the integral?

how about this one:

$$y=\frac{x^5}{6}+\frac{1}{10x^3} \ \ \ \ 1 \leq x \leq 2$$

I found the derivative to be $$\frac{5x^4}{6}-\frac{3}{10x^4}$$..im kinda stuck here..should I just keep going and will I eventually see a substitution?

Keep going! Sometimes, in integration problems (and all Mathematics problems in general), you just have to try.

 

[suspenc3]

ahh..i got them both, they weren't too hard.

 

[benorin]

expand out $$1+\left( \frac{5x^4}{6}-\frac{3}{10x^4}\right) ^{2}$$ and get a common denominator, does the square root go away now?

 

[suspenc3]

Yea I got it..Now this one:

$$y=cos2x$$
$$\frac{dy}{dx} = -2sin2x$$
$$S=2 \pi \int_0^{\pi /6} cos2x \sqrt{1+4sin^22x}dx$$

does this look right so far?

 

[benorin]

yep, keep going.