The work done by
is
![\displaystyle\int_C\vec F\cdot\mathrm d\vec r](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_C%5Cvec%20F%5Ccdot%5Cmathrm%20d%5Cvec%20r)
where
is the given curve and
is the given parameterization of
. We have
![\mathrm d\vec r=\dfrac{\mathrm d\vec r}{\mathrm dt}\mathrm dt=k(1-2t)\,\vec\imath+\vec\jmath](https://tex.z-dn.net/?f=%5Cmathrm%20d%5Cvec%20r%3D%5Cdfrac%7B%5Cmathrm%20d%5Cvec%20r%7D%7B%5Cmathrm%20dt%7D%5Cmathrm%20dt%3Dk%281-2t%29%5C%2C%5Cvec%5Cimath%2B%5Cvec%5Cjmath)
Then the work done by
is
![\displaystyle\int_0^1((t-kt(1-t))\,\vec\imath+kt^2(1-t)\,\vec\jmath)\cdot(k(1-2t)\,\vec\imath+\vec\jmath)\,\mathrm dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E1%28%28t-kt%281-t%29%29%5C%2C%5Cvec%5Cimath%2Bkt%5E2%281-t%29%5C%2C%5Cvec%5Cjmath%29%5Ccdot%28k%281-2t%29%5C%2C%5Cvec%5Cimath%2B%5Cvec%5Cjmath%29%5C%2C%5Cmathrm%20dt)
![=\displaystyle\int_0^1((k-k^2)t-(k-3k^2)t^2-(k+2k^2)t^3)\,\mathrm dt=-\frac k{12}](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E1%28%28k-k%5E2%29t-%28k-3k%5E2%29t%5E2-%28k%2B2k%5E2%29t%5E3%29%5C%2C%5Cmathrm%20dt%3D-%5Cfrac%20k%7B12%7D)
In order for the work to be 1, we need to have
.
Answer:Yes
Step-by-step explanation:You are multiplying the ratios by 4
Answer:
Step-by-step explanation:
It would be more obvious if it says z is a standard Normal variable and that the question is related to statistics.
From "Computing Probabilities Using the Cumulative Table" , the probability of z is less than 1.16, P(z<1.16) = 0.8770.
Answer:
x = 1, x = -1
Step-by-step explanation:
-4x² = -4
x² = 1
x = ![\sqrt{1}](https://tex.z-dn.net/?f=%5Csqrt%7B1%7D)
x = ±1
Answer:
518
Step-by-step explanation:
P = 2
Q = 8
p3 + 64q
Multiply "p" (which is 2) with 3
Multiply "q" (which is 8) with 8
Add the two numbers you get
And you have you answer
I hope this helps. :)