Each square on the grid represents 1 ft2.
2 answers:
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I'm reading this as
with
.
The value of the integral will be independent of the path if we can find a function
that satisfies the gradient equation above.
You have
Integrate
with respect to 
. You get
Differentiate with respect to
. You get
![\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20y%7D%3D%5Cdfrac%7B%5Cpartial%7D%7B%5Cpartial%20y%7D%5Bx%5E2e%5E%7B-y%7D%2Bg%28y%29%5D)
Integrate both sides with respect to to arrive at
So you have
The gradient is continuous for all
, so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is
with
The value of the integral will be independent of the path if we can find a function
You have
Integrate
Differentiate with respect to
Integrate both sides with respect to
So you have
The gradient is continuous for all
A) T(m) = -0.16m +331.3
B) T(2015) = 8.9 . . . seconds
_____
The linear regression function of a graphing calculator can find the equation of the line easily. Or, you can use the 2-point form of the equation of a line.
.. T(m) = (10.5 -14.5)/(2005 -1980)*(m -1980) +14.5
.. T(m) = -0.16(m -1980) +14.5
B) T(2015) = 8.9 . . . seconds
_____
The linear regression function of a graphing calculator can find the equation of the line easily. Or, you can use the 2-point form of the equation of a line.
.. T(m) = (10.5 -14.5)/(2005 -1980)*(m -1980) +14.5
.. T(m) = -0.16(m -1980) +14.5