Find r'(t)·r''(t) of r(t) = <2e²t, 4e⁻²t, te²t>

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SUMMARY

The discussion focuses on calculating the dot product of the first and second derivatives of the vector function r(t) = ⟨2e²ᵗ, 4e⁻²ᵗ, te²ᵗ⟩. The correct derivatives are r'(t) = ⟨4e²ᵗ, -8e⁻²ᵗ, 2te²ᵗ + e²ᵗ⟩ and r''(t) = ⟨8e²ᵗ, 16e⁻²ᵗ, 4te²ᵗ + 4e²ᵗ⟩. The error in the attempted solution was identified as a miscalculation in combining the exponential terms, particularly in the second component of r''(t), which should be 16e⁻²ᵗ instead of 16e²ᵗ.

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Homework Statement


If r(t) = \left\langle\2e^{2t},4e^{-2t},te^{2t} \right\rangle, find r'(t) \cdot r''(t).

Homework Equations


r'(t) = \left\langle4e^{2t},-8e^{-2t},2te^{2t}+e^{2t} \right\rangle
r''(t) = \left\langle8e^{2t},16e^{2t},4te^{2t}+4e^{2t} \right\rangle

The Attempt at a Solution


r'(t) \cdot r''(t) = 32e^{2t} - 128 + 8t^2e^{2t} + 8te^{2t} + 4te^{2t} + 4e^{2t}
r'(t) \cdot r''(t) = 36e^{4t}+8t^2e^{2t}+12te^{2t}-128

I cannot see why my answer is wrong. Please help!
 
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Fixed your post to make it readable:
chickyroger said:

Homework Statement


If \mathbf{r}(t) = \langle 2e^{2t}, 4e^{-2t}, te^{2t} \rangle, find \mathbf{r}&#039;(t)\cdot\mathbf{r}&#039;&#039;(t).

Homework Equations


\mathbf{r}&#039;(t) = \langle 4e^{2t},-8e^{-2t},2te^{2t}+e^{2t} \rangle
\mathbf{r}&#039;&#039;(t) = \langle 8e^{2t},16e^{2t},4te^{2t}+4e^{2t} \rangle

The Attempt at a Solution


\mathbf{r}&#039;(t) \cdot \mathbf{r}&#039;&#039;(t) = 32e^{2t} - 128 + 8t^2e^{2t} + 8te^{2t} + 4te^{2t} + 4e^{2t}
\mathbf{r}&#039;(t) \cdot \mathbf{r}&#039;&#039;(t) = 36e^{4t}+8t^2e^{2t}+12te^{2t}-128

I cannot see why my answer is wrong. Please help!
You didn't combine the exponentials together correctly.
 
chickyroger said:

Homework Statement


If r(t) = \left\langle\2e^{2t},4e^{-2t},te^{2t} \right\rangle, find r'(t) \cdot r''(t).

Homework Equations


r'(t) = \left\langle4e^{2t},-8e^{-2t},2te^{2t}+e^{2t} \right\rangle
This is correct.
r''(t) = \left\langle8e^{2t},16e^{2t},4te^{2t}+4e^{2t} \right\rangle[/quote]
This is not but probably just a typo: the second component is 16e^{-2t}.

The Attempt at a Solution


r'(t) \cdot r''(t) = 32e^{2t} - 128 + 8t^2e^{2t} + 8te^{2t} + 4te^{2t} + 4e^{2t}
And you have propogated the typo: the product in the second term is (-8e^{-2t})(16e^{-2t})= -128e^{-2t}[/tex]<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> <b>r&#039;</b>(t) \cdot <b>r&#039;&#039;</b>(t) = 36e^{4t}+8t^2e^{2t}+12te^{2t}-128<br /> <br /> I cannot see why my answer is wrong. Please help! </div> </div> </blockquote>
 

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