amy0799
  • amy0799
Find k such that the line y = 2x + 18 is tangent to the graph of the function below. y=k√x k=?
Mathematics
katieb
  • katieb
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Loser66
  • Loser66
I told you, take derivative of the curve \(y = k\sqrt x\), what do you get?
amy0799
  • amy0799
1/2k^-1/2
Loser66
  • Loser66
|dw:1440973881338:dw|

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freckles
  • freckles
|dw:1440974118718:dw| you want y=2x+8 and y=ksqrt(x) to have a common point (a,f(a)) and you also want the derivative of 2x+8 to be equal to the derivative of ksqrt(x) at (a,f(a))
amy0799
  • amy0799
right, I understand that, I just don't know how to find it.
freckles
  • freckles
you have 2a+8=k sqrt(a) <--since we want 2a+8 to be equal to k sqrt(a) and you also have 2=k/(2sqrt(a))
Loser66
  • Loser66
Thanks @freckles. Shame on me, I guided him wrong. @amy0799 I am sorry.
freckles
  • freckles
it is just a system of equations to solve
amy0799
  • amy0799
\[\frac{ 2x+8 }{ \sqrt{x}} = k\]
amy0799
  • amy0799
what do I do next?
freckles
  • freckles
\[2a+8=k \sqrt{a} \\ 2=\frac{k}{2 \sqrt{a}} \implies \sqrt{a}=\frac{k}{4} \implies a=\frac{k^2}{16} \text{ insert into first equation }\]
freckles
  • freckles
doing that will give you an equation just in terms of k
amy0799
  • amy0799
insert it into which equation?
freckles
  • freckles
the First equation
freckles
  • freckles
the first equation I wrote above the second equation
freckles
  • freckles
\[2a+8=k \sqrt{a} \\ 2=\frac{k}{2 \sqrt{a}} \implies \sqrt{a}=\frac{k}{4} \implies a=\frac{k^2}{16} \text{ insert into first equation } \] there are equations I'm playing with here the first equation which is on the first line and the second equation which on the second line the second line I used the second equation to express sqrt(a) in terms o k and I also expressed a in terms of k using the second equation we can now write the first equation which I put on the first line in terms of k
amy0799
  • amy0799
so I plug a=k^2/16 into 2a+8=ksqrt(a)?
freckles
  • freckles
yes you also plug in k/4 for sqrt(a)
amy0799
  • amy0799
\[2(\frac{ k^{2} }{ 16 })+8=k \sqrt{k \sqrt{a}}\] this doesn't make sense
freckles
  • freckles
you are write that equation doesn't make sense you should have gotten something else entirely different
freckles
  • freckles
\[2a+8=k \sqrt{a} \\ \text{ we determined } \sqrt{a}=\frac{k}{4} \text{ then we squared both sides \to get } a=\frac{k^2}{16} \\ \text{ plug \in both of these results we obtained from the second equation } \\ \text{ into the first } \\ 2(\frac{k^2}{16})+8=k(\frac{k}{4})\]
freckles
  • freckles
now solve the equation for k (you will see you have to discard a value that you get for k when solving this equation because of the equation sqrt(a)=k/4 which tells us that sqrt(a) is positive and so k has to be positive )
freckles
  • freckles
do you know how to solve the above quadratic
amy0799
  • amy0799
not really
freckles
  • freckles
k*k=k^2 and 2/16=1/8 you have: \[\frac{k^2}{8}+8=\frac{k^2}{4} \\ \text{ you can multiply both sides by 8 \to get } \\ k^2+64=2k^2 \] can you solve from here?
amy0799
  • amy0799
k=8?
freckles
  • freckles
\[64=k^2 \\ k=8 \text{ or } -8 \\ \text{ but we did mentioned that } k>0 \\ \text{ so yes } k=8\]
amy0799
  • amy0799
thank you! can u help me with another one?
freckles
  • freckles
i can try yes
freckles
  • freckles
but
freckles
  • freckles
before we go on you need to understand this one
freckles
  • freckles
like do you have any questions? do you understand how I solved the system of equations?
amy0799
  • amy0799
yea I understood that
amy0799
  • amy0799
Consider the following function. f(t) = 2t^2 − 3 (a) Find the average rate of change of the function below over the interval [3, 3.1]. (b) Compare this average rate of change with the instantaneous rates of change at the endpoints of the interval. (at t = 3) (at t = 3.1)
freckles
  • freckles
\[\text{average rate of change of } f \text{ from }t=a \text{ to } t=b \\ \text{ is given by the formula } \frac{f(b)-f(a)}{b-a}\]
freckles
  • freckles
\[\text{ instantaneous rate of change of} f \text{ at } x=c \text{ is given by } f'(c)\]
amy0799
  • amy0799
so what's the first step I need to do?
freckles
  • freckles
you used the formula above to find the average rate of change from t=3 to t=3.1...
freckles
  • freckles
replace b with 3.1 and a with 3
amy0799
  • amy0799
\[\frac{ f(3.1)-f(3) }{ 3.1-3 }\]
freckles
  • freckles
yes now evaluate f(3.1) and f(3) and do the difference on top and bottom
freckles
  • freckles
and then the division of the top by the bottom
amy0799
  • amy0799
I don't understand what you mean
freckles
  • freckles
f(t) is given as 2t^2-3 you are to find f(3.1) and f(3) by using f(t)=2t^2-3 replace t with 3.1 and get a result for f(3.1) replace t with 3 and get a result for f(3)
amy0799
  • amy0799
f(3.1)=16.22 f(3)=15
freckles
  • freckles
\[\frac{f(3.1)-f(3)}{3.1-3}=\frac{16.22-15}{3.1-3}\] now find the difference for both top and bottom
freckles
  • freckles
the difference of 16.22 and 15 is another way to say 16.22-15 if you are confused by the word difference
amy0799
  • amy0799
12.2
amy0799
  • amy0799
is that the average rate of change?
freckles
  • freckles
yes that is the average rate of change of f(t)=2t^2-3 from t=3 to t=3.1
amy0799
  • amy0799
I got part b wrong
freckles
  • freckles
did you find the f'(3) and f'(3.1)
amy0799
  • amy0799
yea isn't it 16.22 and 15?
freckles
  • freckles
no
freckles
  • freckles
f' means to find the derivative of f
freckles
  • freckles
and you just plugged into original function
freckles
  • freckles
did you not look earlier at how I said to find the instantaneous rate of change?
freckles
  • freckles
I posted the following two comments: \[\text{average rate of change of } f \text{ from }t=a \text{ to } t=b \\ \text{ is given by the formula } \frac{f(b)-f(a)}{b-a}\] \[\text{ instantaneous rate of change of} f \text{ at } x=c \text{ is given by } f'(c)\]
amy0799
  • amy0799
I don't understand it
freckles
  • freckles
you don't know how to find the derivative of 2t^2-3?
freckles
  • freckles
do you know power rule and constant rule? if not use the definition of derivative
amy0799
  • amy0799
4t.
freckles
  • freckles
yes so what do you not understand?
freckles
  • freckles
you have f'(t)=4t and you are asked to evaluate f'(t) at the endpoints
freckles
  • freckles
the endpoints are 3 and 3.1
freckles
  • freckles
f'(3)=? f'(3.1)=?
amy0799
  • amy0799
12 and 12.4
amy0799
  • amy0799
my very first question. I got k=8 wrong
freckles
  • freckles
did you look at what we wrote and what you asked?
freckles
  • freckles
you should see we changed 18 to 8
freckles
  • freckles
you know the line being y=2x+18 instead of y=2x+8
amy0799
  • amy0799
is the answer not 8?
freckles
  • freckles
the answer is k=8 if the question involved y=2x+8 but the equation was y=2x+18 please look at our work and your question our answer was for if the line was y=2x+8 which I started over and over but the equation was y=2x+18 so just redo the work for the equation being y=2x+18 instead of it being y=2x+8
freckles
  • freckles
which I stated over and over in the answer*
freckles
  • freckles
do you understand?
freckles
  • freckles
the question was: Find k such that the line y = 2x + 18 is tangent to the graph of the function below. y=k√x k=? however our work was for: Find k such that the line y = 2x + 8 is tangent to the graph of the function below. y=k√x k=?
freckles
  • freckles
you should be able to do your question now if you actually understand the work we did to find k for this other question
amy0799
  • amy0799
can you help me with it? @freckles
freckles
  • freckles
I thought you said you understand the process?
freckles
  • freckles
\[\text{ solve the system of equations } \\ (2x+18)'|_{x=a}=(k \sqrt{x})'|_{x=a} \\ 2a+18=k \sqrt{a}\]
freckles
  • freckles
it is the same process we did above
freckles
  • freckles
except there is just a little number change
amy0799
  • amy0799
I understood how you solved the quadratic, but not how you got a
freckles
  • freckles
\[2a+8=k \sqrt{a} \\ 2=\frac{k}{2 \sqrt{a}} \implies \sqrt{a}=\frac{k}{4} \implies a=\frac{k^2}{16} \text{ insert into first equation } \] is this what you are talking about? you don't know how I solved the second equation for sqrt(a)?
freckles
  • freckles
\[2=\frac{k}{2 \sqrt{a}} \\ \text{ I just multiplied } 2 \sqrt{a} \text{ on both sides } 2(2 \sqrt{a})=k \\ 4 \sqrt{a}=k \\ \text{ and then divided both sides by } 4\]
freckles
  • freckles
and the only different and this set of equations is 8 is supposed to be 18
freckles
  • freckles
\[2a+18=k \sqrt{a} \\ \sqrt{a}=\frac{k}{4} \text{ squaring both sides gives } a=\frac{k^2}{16} \text{ replace the items in the first equation }\]
amy0799
  • amy0799
k=18?
freckles
  • freckles
how did you get that?
freckles
  • freckles
did you replace a with k^2/16 and sqrt(a) with k/4?
freckles
  • freckles
\[2a+18=k \sqrt{a} \\ 2(\frac{k^2}{16})+18=k(\frac{k}{4})\]
freckles
  • freckles
solve the equation for k
amy0799
  • amy0799
\[\frac{ 18k^{2} }{ 8 }+324=\frac{ 18k ^{2} }{ 4 }\] is this right so far?
freckles
  • freckles
why did you choose to multiply both sides by 18
amy0799
  • amy0799
that's how u did it before
freckles
  • freckles
no I multiplied both sides by something to clear the fractions not to make the problem more fractionie (I know that isn't a word) \[\frac{2}{16}=\frac{1}{8} \\ k \cdot k=k^2 \\ \text{ you have } \\ 2(\frac{k^2}{16})+18=k(\frac{k}{4}) \\ \text{ so this is equivlant to } \\ \frac{k^2}{8}+18=\frac{k^2}{4} \\ \text{ if you multiply both sides by 8 } \text{ you can clear the fractions }\]
freckles
  • freckles
the denominators were same in previous previous problem and I chose to multiply both sides by 8 then so it shouldn't change here
freckles
  • freckles
unless you want to make the problem more complicated looking :p
amy0799
  • amy0799
k=12
freckles
  • freckles
yes \[k^2+8(18)=2k^2 \\ 8(18)=k^2 \\ 16(9)=k^2 \\ 4(3)=k \\ k=12\] and again we chose 12 instead of -12 because we wanted k to be positive @amy0799 you might want to go and review how to solve algebraic equations practice, practice and you can get better at this I promise
amy0799
  • amy0799
yeaa I know >.< I have one more I need help with
freckles
  • freckles
you can post it but brb
freckles
  • freckles
ok i'm here
amy0799
  • amy0799
Consider the following function. \[f(x)=\sqrt{9-x^{2}}\] Find the derivative from the left at x = 3. If it does not exist, enter NONE. Find the derivative from the right at x = 3. If it does not exist, enter NONE.
freckles
  • freckles
do you know what the function looks like?
freckles
  • freckles
|dw:1440979836143:dw| does the function exist to the right of x=3?
amy0799
  • amy0799
no
freckles
  • freckles
so we already know one answer is does not exist... now let's look at the left side: \[\text{ \left derivative of } f \text{ at } x=3 \text{ is given by } \lim_{x \rightarrow 3^-}\frac{f(x)-f(3)}{x-3} \\ \]
amy0799
  • amy0799
it doesn't exist either
freckles
  • freckles
correct
amy0799
  • amy0799
thank you so so much and I'm really sorry if I took up ur time!!
freckles
  • freckles
It is only fine if you take up my time if you have learned something
freckles
  • freckles
do you feel like you learned anything
amy0799
  • amy0799
I learned a lot. thank you!
freckles
  • freckles
np :)

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