Exponential Growth And Composite Functions

[ardentmed]

Hey guys,

Few more questions for the problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

Question:

The first one starts off easy but I found that it gets progress more challenging later on. So the rate of spread should be p'(t) which is:

p'(t) = 5/[e^.5t * (1+10e^(-.5t))^2]As for 1b, I just used chain rule and product rule together to get:

f''(x) = 6xy'(x^2) + 4(x^3)g''(x^2)And finally, for the second question, if the tangent it horizontal, then dy/dx = 0, right?

Therefore, you solve for x, which leads you to:

sinx=-1
x=arcsin(-1)
x= -π/2Which leads to ( -π/2 , 0 ) as the co-ordinates after substituting x = -π/2 into the original function.

Thanks in advance.

 

[zzephod]

Hey guys, ...And finally, for the second question, if the tangent it horizontal, then dy/dx = 0, right?

Therefore, you solve for x, which leads you to:

sinx=-1
x=arcsin(-1)
x= -π/2Which leads to ( -π/2 , 0 ) as the co-ordinates after substituting x = -π/2 into the original function.

Thanks in advance.

For the second question:

$$y'=-\frac{\mathrm{sin}\left( x\right) }{\mathrm{sin}\left( x\right) +2}-\frac{{\mathrm{cos}\left( x\right) }^{2}}{{\left( \mathrm{sin}\left( x\right) +2\right) }^{2}}$$

so: $y'=0$ means:

$$\begin{aligned}
-\frac{\mathrm{sin}\left( x\right) }{2+\mathrm{sin}\left( x\right) }-\frac{{\mathrm{cos}\left( x\right) }^{2}}{{\left( 2+ \mathrm{sin}\left( x\right) \right) }^{2}}&=0,\\
\sin(x)(2+\sin(x))+(\cos(x))^2&=0,\\
2\sin(x)+1&=0,\\
\sin(x)&=-1/2 \dots
\end{aligned} $$

 

[ardentmed]

Horizontal Tangent Lines

Question 1: =Under certain circumstances a rumor spreads according to the equation p(t) = 1/(1+10e^(−0.5t)) ,
where p(t) is the proportion of the population that knows the rumor at time t.
constants.
Find the rate of spread of the rumor.

Question 2: Find the coordinates for the points on the curve where y= cos(x)/(2+sin(x)) $0\le x\le 2\pi $ is horizontal. The first one starts off easy but I found that it gets progress more challenging later on. So the rate of spread should be p'(t) which is:

p'(t) = 5/[e^.5t * (1+10e^(-.5t))^2]As for 1b, I just used chain rule and product rule together to get:

f''(x) = 6xy'(x^2) + 4(x^3)g''(x^2)And finally, for the second question, if the tangent it horizontal, then dy/dx = 0, right?

Therefore, you solve for x, which leads you to:

sinx=-1
x=arcsin(-1)
x= -π/2Which leads to ( -π/2 , 0 ) as the co-ordinates after substituting x = -π/2 into the original function.

Thanks in advance.

 

[ardentmed]

For the second question:

$$y'=-\frac{\mathrm{sin}\left( x\right) }{\mathrm{sin}\left( x\right) +2}-\frac{{\mathrm{cos}\left( x\right) }^{2}}{{\left( \mathrm{sin}\left( x\right) +2\right) }^{2}}$$

so: $y'=0$ means:

$$\begin{aligned}
-\frac{\mathrm{sin}\left( x\right) }{2+\mathrm{sin}\left( x\right) }-\frac{{\mathrm{cos}\left( x\right) }^{2}}{{\left( 2+ \mathrm{sin}\left( x\right) \right) }^{2}}&=0,\\
\sin(x)(2+\sin(x))+(\cos(x))^2&=0,\\
2\sin(x)+1&=0,\\
\sin(x)&=-1/2 \dots
\end{aligned} $$

Alright, so with that in mind, I re-did the question and computed:

arcsin(-1/2) = x

Thus,

x= 11$\pi$ /6, 7$\pi$ / 6.

Giving the points

(11$\pi$ /6, 2/√3) and (7$\pi$ / 6 , -2/√3)

 

[MarkFL]

I agree with your derivation of $p'(t)$, but do not see question 1b).

For the second question, yes, we want to require \(\displaystyle \d{y}{x}=0\). But, you have not found the correct condition, and since you have not shown your work, I have no idea where you went wrong. Please show your work in computing the derivative.

 

[MarkFL]

I have merged two threads together since they are the same questions. Do you see how double posting lead to me duplicating the efforts of zzephod? I value my time, and don't want to see it wasted, nor do I want to see the time of others wasted as well.

Please do not create a new thread dealing with questions you have already posted.

 

[ardentmed]

I agree with your derivation of $p'(t)$, but do not see question 1b).

For the second question, yes, we want to require \(\displaystyle \d{y}{x}=0\). But, you have not found the correct condition, and since you have not shown your work, I have no idea where you went wrong. Please show your work in computing the derivative.

My apologies for the oversight.

Regardless, I did it again and came up with the same answer:

arcsin(-1/2) = x

Therefore, this value can be computed by drawing out the 45 degree triangle with 1-1-√2 sides. Therefore, the solutions should be 2π - π/4 = 11π /6, and π + π/4 = 7π / 6

Thus,

x= 11π /6, 7π / 6.

Giving the points

(11π /6, 2/√3) and (7π / 6 , -2/√3)