## amy0799 one year ago Find k such that the line y = 2x + 18 is tangent to the graph of the function below. y=k√x k=?

1. Loser66

I told you, take derivative of the curve $$y = k\sqrt x$$, what do you get?

2. amy0799

1/2k^-1/2

3. Loser66

|dw:1440973881338:dw|

4. 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))

5. amy0799

right, I understand that, I just don't know how to find it.

6. 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))

7. Loser66

Thanks @freckles. Shame on me, I guided him wrong. @amy0799 I am sorry.

8. freckles

it is just a system of equations to solve

9. amy0799

$\frac{ 2x+8 }{ \sqrt{x}} = k$

10. amy0799

what do I do next?

11. freckles

$2a+8=k \sqrt{a} \\ 2=\frac{k}{2 \sqrt{a}} \implies \sqrt{a}=\frac{k}{4} \implies a=\frac{k^2}{16} \text{ insert into first equation }$

12. freckles

doing that will give you an equation just in terms of k

13. amy0799

insert it into which equation?

14. freckles

the First equation

15. freckles

the first equation I wrote above the second equation

16. 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

17. amy0799

so I plug a=k^2/16 into 2a+8=ksqrt(a)?

18. freckles

yes you also plug in k/4 for sqrt(a)

19. amy0799

$2(\frac{ k^{2} }{ 16 })+8=k \sqrt{k \sqrt{a}}$ this doesn't make sense

20. freckles

you are write that equation doesn't make sense you should have gotten something else entirely different

21. freckles

$2a+8=k \sqrt{a} \\ \text{ we determined } \sqrt{a}=\frac{k}{4} \text{ then we squared both sides \to get } a=\frac{k^2}{16} \\ \text{ plug \in both of these results we obtained from the second equation } \\ \text{ into the first } \\ 2(\frac{k^2}{16})+8=k(\frac{k}{4})$

22. 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 )

23. freckles

do you know how to solve the above quadratic

24. amy0799

not really

25. 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?

26. amy0799

k=8?

27. freckles

$64=k^2 \\ k=8 \text{ or } -8 \\ \text{ but we did mentioned that } k>0 \\ \text{ so yes } k=8$

28. amy0799

thank you! can u help me with another one?

29. freckles

i can try yes

30. freckles

but

31. freckles

before we go on you need to understand this one

32. freckles

like do you have any questions? do you understand how I solved the system of equations?

33. amy0799

yea I understood that

34. 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)

35. freckles

$\text{average rate of change of } f \text{ from }t=a \text{ to } t=b \\ \text{ is given by the formula } \frac{f(b)-f(a)}{b-a}$

36. freckles

$\text{ instantaneous rate of change of} f \text{ at } x=c \text{ is given by } f'(c)$

37. amy0799

so what's the first step I need to do?

38. freckles

you used the formula above to find the average rate of change from t=3 to t=3.1...

39. freckles

replace b with 3.1 and a with 3

40. amy0799

$\frac{ f(3.1)-f(3) }{ 3.1-3 }$

41. freckles

yes now evaluate f(3.1) and f(3) and do the difference on top and bottom

42. freckles

and then the division of the top by the bottom

43. amy0799

I don't understand what you mean

44. 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)

45. amy0799

f(3.1)=16.22 f(3)=15

46. 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

47. freckles

the difference of 16.22 and 15 is another way to say 16.22-15 if you are confused by the word difference

48. amy0799

12.2

49. amy0799

is that the average rate of change?

50. freckles

yes that is the average rate of change of f(t)=2t^2-3 from t=3 to t=3.1

51. amy0799

I got part b wrong

52. freckles

did you find the f'(3) and f'(3.1)

53. amy0799

yea isn't it 16.22 and 15?

54. freckles

no

55. freckles

f' means to find the derivative of f

56. freckles

and you just plugged into original function

57. freckles

did you not look earlier at how I said to find the instantaneous rate of change?

58. freckles

I posted the following two comments: $\text{average rate of change of } f \text{ from }t=a \text{ to } t=b \\ \text{ is given by the formula } \frac{f(b)-f(a)}{b-a}$ $\text{ instantaneous rate of change of} f \text{ at } x=c \text{ is given by } f'(c)$

59. amy0799

I don't understand it

60. freckles

you don't know how to find the derivative of 2t^2-3?

61. freckles

do you know power rule and constant rule? if not use the definition of derivative

62. amy0799

4t.

63. freckles

yes so what do you not understand?

64. freckles

you have f'(t)=4t and you are asked to evaluate f'(t) at the endpoints

65. freckles

the endpoints are 3 and 3.1

66. freckles

f'(3)=? f'(3.1)=?

67. amy0799

12 and 12.4

68. amy0799

my very first question. I got k=8 wrong

69. freckles

did you look at what we wrote and what you asked?

70. freckles

you should see we changed 18 to 8

71. freckles

you know the line being y=2x+18 instead of y=2x+8

72. amy0799

73. 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

74. freckles

which I stated over and over in the answer*

75. freckles

do you understand?

76. 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=?

77. 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

78. amy0799

can you help me with it? @freckles

79. freckles

I thought you said you understand the process?

80. freckles

$\text{ solve the system of equations } \\ (2x+18)'|_{x=a}=(k \sqrt{x})'|_{x=a} \\ 2a+18=k \sqrt{a}$

81. freckles

it is the same process we did above

82. freckles

except there is just a little number change

83. amy0799

I understood how you solved the quadratic, but not how you got a

84. 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)?

85. freckles

$2=\frac{k}{2 \sqrt{a}} \\ \text{ I just multiplied } 2 \sqrt{a} \text{ on both sides } 2(2 \sqrt{a})=k \\ 4 \sqrt{a}=k \\ \text{ and then divided both sides by } 4$

86. freckles

and the only different and this set of equations is 8 is supposed to be 18

87. freckles

$2a+18=k \sqrt{a} \\ \sqrt{a}=\frac{k}{4} \text{ squaring both sides gives } a=\frac{k^2}{16} \text{ replace the items in the first equation }$

88. amy0799

k=18?

89. freckles

how did you get that?

90. freckles

did you replace a with k^2/16 and sqrt(a) with k/4?

91. freckles

$2a+18=k \sqrt{a} \\ 2(\frac{k^2}{16})+18=k(\frac{k}{4})$

92. freckles

solve the equation for k

93. amy0799

$\frac{ 18k^{2} }{ 8 }+324=\frac{ 18k ^{2} }{ 4 }$ is this right so far?

94. freckles

why did you choose to multiply both sides by 18

95. amy0799

that's how u did it before

96. 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) $\frac{2}{16}=\frac{1}{8} \\ k \cdot k=k^2 \\ \text{ you have } \\ 2(\frac{k^2}{16})+18=k(\frac{k}{4}) \\ \text{ so this is equivlant to } \\ \frac{k^2}{8}+18=\frac{k^2}{4} \\ \text{ if you multiply both sides by 8 } \text{ you can clear the fractions }$

97. 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

98. freckles

unless you want to make the problem more complicated looking :p

99. amy0799

k=12

100. 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

101. amy0799

yeaa I know >.< I have one more I need help with

102. freckles

you can post it but brb

103. freckles

ok i'm here

104. 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.

105. freckles

do you know what the function looks like?

106. freckles

|dw:1440979836143:dw| does the function exist to the right of x=3?

107. amy0799

no

108. freckles

so we already know one answer is does not exist... now let's look at the left side: $\text{ \left derivative of } f \text{ at } x=3 \text{ is given by } \lim_{x \rightarrow 3^-}\frac{f(x)-f(3)}{x-3} \\$

109. amy0799

it doesn't exist either

110. freckles

correct

111. amy0799

thank you so so much and I'm really sorry if I took up ur time!!

112. freckles

It is only fine if you take up my time if you have learned something

113. freckles

do you feel like you learned anything

114. amy0799

I learned a lot. thank you!

115. freckles

np :)