fiora

A spherical ballon is being inflated in such a way that its radius is increasing at the constant rate of 3cm/min.\(\Rightarrow \frac{dr}{dt}=3cm/min\)

The relation between the volume and radius of the spherical ballon is \(v(r)=4*{r}^{3}/3\)

The dervative of volume v  respect to variable v is \(\frac{dv}{dr}=4{r}^{2}\)

At time 5 minutes \(\frac{dv}{dr}=4*5^2=100\)cm^2

Applying chain rule,\(\frac{dv}{dt}=\frac{dv}{dr}*\frac{dr}{dt}\)

\(\frac{dv}{dt}=100*cm^2*3cm/min=300cm^3/min\)

This is a kind of problem relative rate of change,application of differvative.

At time equal to 0 years(t=0),the original isotope has 85

when dropped to 1/6,the isotope has 85/6

substitute it into M(t) and we have

\(85/6=85*e^\left (-t/719 \right )\)  

\(1/6=e^\left (-t/719 \right )\)

ln(1/6)=-t/719

-ln(6)=-t/719

ln6=t/719

t=719*ln6

t=ln(6^719)=1*ln(6^719)

6^719 about equal to infinity

Thank you,asinus!Nice explanation and graphics representations. 

x=t+1     t=x-1  \(\frac{dt}{dx}=1\)

y=3t+2    \(\frac{dy}{dt}=3\)

applying chain rule we have

\(\frac{dy}{dx}=\frac{dy}{dt}*\frac{dt}{dx}=3\)

 the average velocity of the object over a time interval from time 4+h seconds to time 4 seconds is 

\(\frac{S(4+h）-S(4)}{(h+4)-4}=\frac{（\sqrt{4+h}-\sqrt{4})}{h}=\frac{（\sqrt{4+h}-\sqrt{4})}{h}*\frac{\sqrt{4+h}+\sqrt{4}}{\sqrt{4+h}+\sqrt{4}}=\frac{h}{h*(\sqrt{4+h}+\sqrt{4})}=1/(\sqrt{4+h}+\sqrt{4})\)

(2)\(\lim_{h\rightarrow 0} 1/(\sqrt{4+h}+\sqrt{4})=1/(\sqrt{4}+\sqrt{4})=1/2\)

Now,I want to show you guys some physics and mathematics.

acceleration is the change in velocity in a given amount of time.\(a=\frac{V-Vo}{t}\)

where V is the final veocity and Vo is the initial velocity.

rearrange,we have \(V=at+Vo\)If the object's accelration is constant,then the graph of the velocity is a linear function.

But I would like to express it in y=mx+b form in my graph.And here is my graph

The area of the triangle is the sum of the  two base times the hight divide by two.or A=(b1+b2)*h/2

In here ,we could express it as \(A=(V+Vo)*\Delta t/2\)

the area under the curve(line) of function of velocity is the displacement.

Therefore,\(\Delta S=(V+Vo)*\Delta t/2\)\(a=\frac{V-Vo}{t}\Rightarrow t=\frac{V-Vo}{a}\)

If the object start moving at time 0 seconds 

\(\Delta t=T-To=t\)

the displacement of the object from time 0 seonds is \(\Delta S=\frac{V^2-Vo^2}{2a}\)

substitute V=at+Vo into \(\Delta S=(V+Vo)*\Delta t/2\)

we have \(\Delta S=(2Vo+at)*t/2\Rightarrow \Delta S=1/2at^2+Vo*t\)

displacement equal final position subract inital position \(\Delta S=S-So\)

\(S=\Delta S+So\)\(S=1/2at^2+Vo*t+So\)

Confirm my previous equation by using integration

\(V=\frac{ds}{dt}=a*t+Vo \Rightarrow \int \frac{ds}{dt}dt=\int at+Vo dt\Rightarrow S=1/2at^2+Vo*t+C\) for some costant C

In here C is the initial position,so \(S=1/2at^2+Vo*t+So\)

The instantaneous velocity is  the derivative of the position function s=f(t) with respect to time.At tiem t,the velocity is 

V(t)=\(\frac{ds}{dt}=\lim_{x\rightarrow \Delta t } \frac{S(t+\Delta t)-S(t)}{\Delta t} \)

(1)v(t)=\(\frac{ds}{dt} t^2=2t\)

(2)\(\lim_{t\rightarrow 2} 2t=2*2=4\)

not good at calculus,might be wrong

In the diagram, DAB = CBA = 90 degrees, EC = 5cm and ED = 10cm. The point F is the reflection of C in the line AB, and DF intersects AB at E. Given that the area of Traingle CED is 12.5 cm squared, and angle DEC =30 degrees, what is the length of CD?

read"the point F is the reflection of C in the line AB, and DF intersects AB at E."

the point F is the reflection of C in the line AB mean point F in the line AB;DF intersects AB at E mean point e is also in line AB (this is not important to this question tho)

"EC = 5cm and ED = 10cm.angle DEC =30"

CD^2=DE^2+CE^2-2DE*CE*cos(angle DEC)

CD^2=100cm^2+25cm^2-100cm*sqrt(3)/2

$${{\mathtt{CD}}}^{{\mathtt{2}}} = \left({\mathtt{125}}{\mathtt{\,-\,}}{\mathtt{50}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}{{\mathtt{cm}}}^{{\mathtt{2}}}$$

$${\mathtt{CD}} = {\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}}}{\mathtt{\,\times\,}}{\mathtt{cm}}$$ 

$${\mathtt{CD}} = {\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}}} \Rightarrow {\mathtt{CD}} = {\mathtt{6.196\: \!568\: \!374\: \!637\: \!379\: \!5}}$$ cm

10 miles per hours =1/6 miles per minute

15 miles per hours=1/4 miles per minute

Now, assume he need x minute to rides  his bicycle to  work.

if he averages 10 miles per hour ,then he need (x+2) minute to arrives at work

if he averages 15 miles per hour,then he need (x-1)minute to arrives at work

distance equal velocity multiply by time. d=v*t

distance did not change,so we can set up an equation like....

1/6*(x+2)=1/4*(x-1)

        x+2=1.5(x-1)

        x+2=1.5x-1.5

       0.5x=3.5

       x=7

If if he average speed is 15  miles per hour (1/4 miles per minute), he need 7-1=6 minute  to arrive at work 

6*(1/4)=3/2 miles=1.5 miles

Same,If he average speed is 10 miles per hour (1/6 miles per minute),he need 7+2=9 minute to arrive at work.

9*(1/6)=3/2 miles=1.5 miles (I just checking)