<span>It led to an intensification of the nuclear arms race. </span>
Answer:
a) The gradient of a function is the vector of partial derivatives. Then
![\nabla f=(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})=(y,x)=y\hat{i} + x\hat{j}](https://tex.z-dn.net/?f=%5Cnabla%20f%3D%28%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20x%7D%2C%20%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20y%7D%29%3D%28y%2Cx%29%3Dy%5Chat%7Bi%7D%20%2B%20x%5Chat%7Bj%7D)
b) It's enough evaluate P in the gradient.
![\nabla f(P)=(-4,-4)=-4\hat{i} - 4 \hat{j}](https://tex.z-dn.net/?f=%5Cnabla%20f%28P%29%3D%28-4%2C-4%29%3D-4%5Chat%7Bi%7D%20-%204%20%5Chat%7Bj%7D)
c) The directional derivative of f at P in direction of V is the dot produtc of
and v.
![\nabla f(P) v=(-4,-4)\left[\begin{array}{ccc}2\\3\end{array}\right] =(-4)2+(-4)3=-20](https://tex.z-dn.net/?f=%5Cnabla%20f%28P%29%20v%3D%28-4%2C-4%29%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5C%5C3%5Cend%7Barray%7D%5Cright%5D%20%3D%28-4%292%2B%28-4%293%3D-20)
d) The maximum rate of change of f at P is the magnitude of the gradient vector at P.
![||\nabla f(P)||=\sqrt{(-4)^2+(-4)^2}=\sqrt{32}=4\sqrt{2}](https://tex.z-dn.net/?f=%7C%7C%5Cnabla%20f%28P%29%7C%7C%3D%5Csqrt%7B%28-4%29%5E2%2B%28-4%29%5E2%7D%3D%5Csqrt%7B32%7D%3D4%5Csqrt%7B2%7D)
e) The maximum rate of change occurs in the direction of the gradient. Then
![v=\frac{1}{4\sqrt{2}}(-4,-4)=(\frac{-1}{\sqrt{2}},\frac{-1}{\sqrt{2}})= \frac{-1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B1%7D%7B4%5Csqrt%7B2%7D%7D%28-4%2C-4%29%3D%28%5Cfrac%7B-1%7D%7B%5Csqrt%7B2%7D%7D%2C%5Cfrac%7B-1%7D%7B%5Csqrt%7B2%7D%7D%29%3D%20%5Cfrac%7B-1%7D%7B%5Csqrt%7B2%7D%7D%5Chat%7Bi%7D-%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%5Chat%7Bj%7D)
is the direction vector in which the maximum rate of change occurs at P.
Using implicit differentiation, it is found that y is changing at a rate of 10 units per second.
---------------------
The equation is:
![y = x^3 + x^2 + 2](https://tex.z-dn.net/?f=y%20%3D%20x%5E3%20%2B%20x%5E2%20%2B%202)
Applying implicit differentiation in function of t, we have that:
![\frac{dy}{dt} = 3x^2\frac{dx}{dt} + 2x\frac{dx}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%20%3D%203x%5E2%5Cfrac%7Bdx%7D%7Bdt%7D%20%2B%202x%5Cfrac%7Bdx%7D%7Bdt%7D)
- x-values changing at a rate of 2 units per second means that
![\frac{dx}{dt} = 2](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bdt%7D%20%3D%202)
- Point Q(1,4) means that
.
We want to find
, thus:
![\frac{dy}{dt} = 3x^2\frac{dx}{dt} + 2x\frac{dx}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%20%3D%203x%5E2%5Cfrac%7Bdx%7D%7Bdt%7D%20%2B%202x%5Cfrac%7Bdx%7D%7Bdt%7D)
![\frac{dy}{dt} = 3(1)^2(2) + 2(1)(2) = 6 + 4 = 10](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%20%3D%203%281%29%5E2%282%29%20%2B%202%281%29%282%29%20%3D%206%20%2B%204%20%3D%2010)
y is changing at a rate of 10 units per second.
A similar problem is given at brainly.com/question/9543179
(8-0)^2+(6-0)^2=10^2
r=10
(x-8)^2+(y-6)^2=100
Answer:
B.
Step-by-step explanation: