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please help!
- Thread starter elise13
- Start date
Hello, elise13!
By now, you've seen that your formatting is not preserved in your post.
(b) We are given \(\displaystyle r\,=\,1.5\) and asked to find \(\displaystyle t: \L\;\frac{1\,+\,2t}{1\,+\,t}\:=\:1.5\)
\(\displaystyle \L\;\;\;1\,+\,2t\:=\:1.5\,+\,1.5t\;\;\Rightarrow\;\;0.5t\:=\:0.5\;\;\Rightarrow\;\;t\,=\,1\) sec.
(c) Area formula: \(\displaystyle \:A\:=\:\pi r^2\)
Differentiate with respect to time: \(\displaystyle \L\frac{dA}{dt}\:=\:2\pi r\left(\frac{dr}{dt}\right)\)
\(\displaystyle \;\;\frac{dr}{dt}\:=\:\frac{(1\,+\,t)\cdot2\,-\,(1\,+\,2t)\cdot1}{(1\,+\,t)^2} \:=\:\frac{1}{(1\,+\,t)^2}\)
\(\displaystyle \;\;\)When \(\displaystyle \,r\,=\,1.5,\:t\,=\,1\;\) (part b) . . . so \(\displaystyle \,\frac{dr}{dt}\,=\,\frac{1}{4}\)
Therefore: \(\displaystyle \L\:\frac{dA}{dt}\:=\:2\pi(1.5)\left(\frac{1}{4}\right)\:=\:\frac{3}{4}\pi\)\(\displaystyle \;m^2/h\;\;\) **
(d) No . . . There are many ways to explain this.
Here's one way . . .
Apply long division to: \(\displaystyle \L\,\frac{2t\,+\,1}{t\,+\,1}\;\;\Rightarrow\;\;2\,-\,\frac{1}{t\,+\,1}\)
The radius is always 2 cm ... minus a fraction.
\(\displaystyle \;\;\)Hence, the radius is always less than 2.
Even if we wait for a year ... 31,536,000 seconds
\(\displaystyle \;\;\)the radius will be: \(\displaystyle \,2\,-\,\frac{1}{31.536.000}\:\approx\;1.999999968\) cm.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**
The first time, I found \(\displaystyle \,\frac{dr}{dt}\) . . . not \(\displaystyle \frac{dA}{dt}\)
Fortunately, two people caught my error and notified me.
Thanks for the heads-up, guys!
By now, you've seen that your formatting is not preserved in your post.
(a) We are asked for \(\displaystyle r\) when \(\displaystyle t\,=\,0:\L\;\;r(0)\:=\:\frac{1\,+\,2\cdot0}{1\,+\,0}\:=\:1\) cm.The radius of a circular juice blot on a piece of paper towel t seconds after is was first seen
is modelled by: \(\displaystyle \,r(t)\:=\:\frac{1\,+\,2t}{1\,+\,t}\) where \(\displaystyle r\) is measured in centimeters.
Calculate:
a. the radius of the blot when it was first observed
b. the time at which the radius of the blot was 1.5 cm
c. the rate of increase of the area of the blot when the radius was 1.5 cm
d. According to this model, will the radius of the blot ever reach 2 cm? Explain your answer.
(b) We are given \(\displaystyle r\,=\,1.5\) and asked to find \(\displaystyle t: \L\;\frac{1\,+\,2t}{1\,+\,t}\:=\:1.5\)
\(\displaystyle \L\;\;\;1\,+\,2t\:=\:1.5\,+\,1.5t\;\;\Rightarrow\;\;0.5t\:=\:0.5\;\;\Rightarrow\;\;t\,=\,1\) sec.
(c) Area formula: \(\displaystyle \:A\:=\:\pi r^2\)
Differentiate with respect to time: \(\displaystyle \L\frac{dA}{dt}\:=\:2\pi r\left(\frac{dr}{dt}\right)\)
\(\displaystyle \;\;\frac{dr}{dt}\:=\:\frac{(1\,+\,t)\cdot2\,-\,(1\,+\,2t)\cdot1}{(1\,+\,t)^2} \:=\:\frac{1}{(1\,+\,t)^2}\)
\(\displaystyle \;\;\)When \(\displaystyle \,r\,=\,1.5,\:t\,=\,1\;\) (part b) . . . so \(\displaystyle \,\frac{dr}{dt}\,=\,\frac{1}{4}\)
Therefore: \(\displaystyle \L\:\frac{dA}{dt}\:=\:2\pi(1.5)\left(\frac{1}{4}\right)\:=\:\frac{3}{4}\pi\)\(\displaystyle \;m^2/h\;\;\) **
(d) No . . . There are many ways to explain this.
Here's one way . . .
Apply long division to: \(\displaystyle \L\,\frac{2t\,+\,1}{t\,+\,1}\;\;\Rightarrow\;\;2\,-\,\frac{1}{t\,+\,1}\)
The radius is always 2 cm ... minus a fraction.
\(\displaystyle \;\;\)Hence, the radius is always less than 2.
Even if we wait for a year ... 31,536,000 seconds
\(\displaystyle \;\;\)the radius will be: \(\displaystyle \,2\,-\,\frac{1}{31.536.000}\:\approx\;1.999999968\) cm.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**
The first time, I found \(\displaystyle \,\frac{dr}{dt}\) . . . not \(\displaystyle \frac{dA}{dt}\)
Fortunately, two people caught my error and notified me.
Thanks for the heads-up, guys!