- #1

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1. Square root of 30 is 0 because its a constant correct?

2. R(t) = 5t^(-3/5) <---- I have tried this problem many times with the definition, but I cant come up with the correct solution and I dont know why!

Thanks

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- Thread starter helpm3pl3ase
- Start date

- #1

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1. Square root of 30 is 0 because its a constant correct?

2. R(t) = 5t^(-3/5) <---- I have tried this problem many times with the definition, but I cant come up with the correct solution and I dont know why!

Thanks

- #2

- 1,235

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1. correct

2. rewrite this as [tex] 5\frac{d}{dt} \frac{1}{t^{\frac{3}{5}}} [/tex].

So: [tex] \frac{1}{\sqrt[5]{(t+h)^{3}}} - \frac{1}{\sqrt[5]{t^{3}}} [/tex]. Subtract the fractions. Then rationalize the numerator. take the limit, and then multiply by 5.

2. rewrite this as [tex] 5\frac{d}{dt} \frac{1}{t^{\frac{3}{5}}} [/tex].

So: [tex] \frac{1}{\sqrt[5]{(t+h)^{3}}} - \frac{1}{\sqrt[5]{t^{3}}} [/tex]. Subtract the fractions. Then rationalize the numerator. take the limit, and then multiply by 5.

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- #3

HallsofIvy

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- #4

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hmm iam still confused.. sorry

not about square root of 30.. about the next problem

not about square root of 30.. about the next problem

- #5

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courtigrad, you surely meant the derivative wrt. t, didn't you? ;)

- #6

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yeah, just a typo. my bad. also i notice people type "courtigrad" instead of "courtrigrad".

- #7

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the trouble is: what do you multiply the fraction by to rationalize the numerator.

[tex] t^{\frac{3}{5}} - (t+h)^{\frac{3}{5}} [/tex] multiplied by something to get [tex] t - (t+h) = -h [/tex]

For example, [tex] t^{\frac{1}{2}} - (t+h)^{\frac{1}{2}}(t^{\frac{1}{2}} + (t+h)^{\frac{1}{2}}) = -h [/tex]

[tex] t^{\frac{3}{5}} - (t+h)^{\frac{3}{5}} [/tex] multiplied by something to get [tex] t - (t+h) = -h [/tex]

For example, [tex] t^{\frac{1}{2}} - (t+h)^{\frac{1}{2}}(t^{\frac{1}{2}} + (t+h)^{\frac{1}{2}}) = -h [/tex]

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- #8

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Alright I got it.. Thanks Guys!

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