Formula for curvature for a well behaved curve y=f(x) is
K(x)= ![\frac{|{y}''|}{[1+{y}'^2]^\frac{3}{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B%7C%7By%7D%27%27%7C%7D%7B%5B1%2B%7By%7D%27%5E2%5D%5E%5Cfrac%7B3%7D%7B2%7D%7D)
The given curve is y=7![e^{x}](https://tex.z-dn.net/?f=e%5E%7Bx%7D)
![{y}''=7e^{x}\\ {y}'=7e^{x}](https://tex.z-dn.net/?f=%7By%7D%27%27%3D7e%5E%7Bx%7D%5C%5C%20%7By%7D%27%3D7e%5E%7Bx%7D)
k(x)=![\frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B7e%5E%7Bx%7D%7D%7B%5B%7B1%2B%287e%5E%7Bx%7D%29%5E2%7D%5D%5E%5Cfrac%7B3%7D%7B2%7D%7D)
![{k(x)}'=\frac{7(e^x)(1+49e^{2x})(49e^{2x}-\frac{1}{2})}{[1+49e^{2x}]^{3}}](https://tex.z-dn.net/?f=%7Bk%28x%29%7D%27%3D%5Cfrac%7B7%28e%5Ex%29%281%2B49e%5E%7B2x%7D%29%2849e%5E%7B2x%7D-%5Cfrac%7B1%7D%7B2%7D%29%7D%7B%5B1%2B49e%5E%7B2x%7D%5D%5E%7B3%7D%7D)
For Maxima or Minima
![{k(x)}'=0](https://tex.z-dn.net/?f=%7Bk%28x%29%7D%27%3D0)
![7(e^x)(1+49e^{2x})(98e^{2x}-1)=0](https://tex.z-dn.net/?f=7%28e%5Ex%29%281%2B49e%5E%7B2x%7D%29%2898e%5E%7B2x%7D-1%29%3D0)
→![e^{x}=0∨ 1+49e^{2x}=0∨98e^{2x}-1=0](https://tex.z-dn.net/?f=e%5E%7Bx%7D%3D0%E2%88%A8%201%2B49e%5E%7B2x%7D%3D0%E2%88%A898e%5E%7B2x%7D-1%3D0)
[not possible ∵there exists no value of x satisfying these equation]
→![98e^{2x}-1=0](https://tex.z-dn.net/?f=98e%5E%7B2x%7D-1%3D0)
Solving this we get
x= ![-\frac{1}{2}\ln{98}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cln%7B98%7D)
As you will evaluate
<0 at x=
So this is the point of Maxima. we get y=7×1/√98=1/√2
(x,y)=[
,1/√2]
k(x)=![\lim_{x\to\infty } \frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}](https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%5Cinfty%20%7D%20%5Cfrac%7B7e%5E%7Bx%7D%7D%7B%5B%7B1%2B%287e%5E%7Bx%7D%29%5E2%7D%5D%5E%5Cfrac%7B3%7D%7B2%7D%7D)
k(x)=![\frac{7}{\infty}](https://tex.z-dn.net/?f=%5Cfrac%7B7%7D%7B%5Cinfty%7D)
k(x)=0
Answer:
ok give me a min
Step-by-step explanation:
Answer:
36 degrees
Step-by-step explanation:
I’m pretty sure that’s it
Step-by-step explanation:
2 1/3
=2*3+1/3
=7/3
soh cah toa
that is the saying I use
we will use cah
c is cos
a is ajacent
h is hypotinuse
to find cos K we need to do
adjacent / hypotenuse
can you do that and write it in the comments