<span>y = sqrt(25-x^2) at point (3,4)
The derivative gives us the slope at 3 to be:
-2x
------------ at x=3: -3/4
2sqrt(25-x^2)
</span><span>so we have a vector that is parallel to the slope of the tangent line is: <4,-3>
</span>
<span>the mag = 5 so; unit tangent = <4/5 , -3/5>
</span>
<span>since perpendicular lines have a -1 product between slopes we get the normal to be...
<3/5,4/5>
</span>
<span>It is <4,-3> because it is rise over run. Rise is y component of vector and run is x component of vector.</span>
The area under a graph is given by the integral of that function, evaluated in the interval of interest:
![\displaystyle \int_{-3}^2 x^2+4\;dx = \left[\dfrac{x^3}{3}+4x\right]_{-3}^2 = \left[\dfrac{2^3}{3}+4\cdot 2\right]-\left[\dfrac{(-3)^3}{3}+4\cdot(-3)\right] = \left[\dfrac{8}{3}+8\right]-\left[-9-12\right] = \dfrac{32}{3}+21 = \dfrac{95}{3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-3%7D%5E2%20x%5E2%2B4%5C%3Bdx%20%3D%20%5Cleft%5B%5Cdfrac%7Bx%5E3%7D%7B3%7D%2B4x%5Cright%5D_%7B-3%7D%5E2%20%3D%20%5Cleft%5B%5Cdfrac%7B2%5E3%7D%7B3%7D%2B4%5Ccdot%202%5Cright%5D-%5Cleft%5B%5Cdfrac%7B%28-3%29%5E3%7D%7B3%7D%2B4%5Ccdot%28-3%29%5Cright%5D%20%3D%20%5Cleft%5B%5Cdfrac%7B8%7D%7B3%7D%2B8%5Cright%5D-%5Cleft%5B-9-12%5Cright%5D%20%3D%20%5Cdfrac%7B32%7D%7B3%7D%2B21%20%3D%20%5Cdfrac%7B95%7D%7B3%7D)
Answer:
test? :/
Step-by-step explanation:
C cumulative property of multiplication