Answer:
Reflection
Step-by-step explanation: It is reflection because it is the same thing just reflected
![f(x,y)=\dfrac{y^3}x](https://tex.z-dn.net/?f=f%28x%2Cy%29%3D%5Cdfrac%7By%5E3%7Dx)
a. The gradient is
![\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\vec\imath+\dfrac{\partial f}{\partial y}\,\vec\jmath](https://tex.z-dn.net/?f=%5Cnabla%20f%28x%2Cy%29%3D%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20x%7D%5C%2C%5Cvec%5Cimath%2B%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20y%7D%5C%2C%5Cvec%5Cjmath)
![\boxed{\nabla f(x,y)=-\dfrac{y^3}{x^2}\,\vec\imath+\dfrac{3y^2}x\,\vec\jmath}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cnabla%20f%28x%2Cy%29%3D-%5Cdfrac%7By%5E3%7D%7Bx%5E2%7D%5C%2C%5Cvec%5Cimath%2B%5Cdfrac%7B3y%5E2%7Dx%5C%2C%5Cvec%5Cjmath%7D)
b. The gradient at point P(1, 2) is
![\boxed{\nabla f(1,2)=-8\,\vec\imath+12\,\vec\jmath}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cnabla%20f%281%2C2%29%3D-8%5C%2C%5Cvec%5Cimath%2B12%5C%2C%5Cvec%5Cjmath%7D)
c. The derivative of
at P in the direction of
is
![D_{\vec u}f(1,2)=\nabla f(1,2)\cdot\dfrac{\vec u}{\|\vec u\|}](https://tex.z-dn.net/?f=D_%7B%5Cvec%20u%7Df%281%2C2%29%3D%5Cnabla%20f%281%2C2%29%5Ccdot%5Cdfrac%7B%5Cvec%20u%7D%7B%5C%7C%5Cvec%20u%5C%7C%7D)
It looks like
![\vec u=\dfrac{13}2\,\vec\imath+5\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20u%3D%5Cdfrac%7B13%7D2%5C%2C%5Cvec%5Cimath%2B5%5C%2C%5Cvec%5Cjmath)
so that
![\|\vec u\|=\sqrt{\left(\dfrac{13}2\right)^2+5^2}=\dfrac{\sqrt{269}}2](https://tex.z-dn.net/?f=%5C%7C%5Cvec%20u%5C%7C%3D%5Csqrt%7B%5Cleft%28%5Cdfrac%7B13%7D2%5Cright%29%5E2%2B5%5E2%7D%3D%5Cdfrac%7B%5Csqrt%7B269%7D%7D2)
Then
![D_{\vec u}f(1,2)=\dfrac{\left(-8\,\vec\imath+12\,\vec\jmath\right)\cdot\left(\frac{13}2\,\vec\imath+5\,\vec\jmath\right)}{\frac{\sqrt{269}}2}](https://tex.z-dn.net/?f=D_%7B%5Cvec%20u%7Df%281%2C2%29%3D%5Cdfrac%7B%5Cleft%28-8%5C%2C%5Cvec%5Cimath%2B12%5C%2C%5Cvec%5Cjmath%5Cright%29%5Ccdot%5Cleft%28%5Cfrac%7B13%7D2%5C%2C%5Cvec%5Cimath%2B5%5C%2C%5Cvec%5Cjmath%5Cright%29%7D%7B%5Cfrac%7B%5Csqrt%7B269%7D%7D2%7D)
![\boxed{D_{\vec u}f(1,2)=\dfrac{16}{\sqrt{269}}}](https://tex.z-dn.net/?f=%5Cboxed%7BD_%7B%5Cvec%20u%7Df%281%2C2%29%3D%5Cdfrac%7B16%7D%7B%5Csqrt%7B269%7D%7D%7D)
Yes. The umbrella is smaller than the suitcase. The suitcase is 81'^2, but the umbrella is 14
The answer to this question is that the trains will meet after 3 hours.
We can work this out by considering that is the closing speed of the two
trains is (50+60=)110 miles per hour, then this must mean that the
combined distance that the trains need to travel before they meet is 330
miles. If the time that is taken to travel 330 miles at 110 miles per
hour, then you simply need to divide 330/110 to find your answer - 3
hours.
The answers to the questions are
- The mathematical model is given as s = −16t2 + 1054
- The height after 4.5 seconds is 730 feet
- the time it would take to strike the ground is 8.11 seconds.
<h3>How to solve for the position of the object</h3>
The mathematical model of this problem would be written as
s = −16t2 + v0t + s0
s0 = 1054
then we would have
s = −16t2 + 1054
b. after 4.5 seconds the height is going to be
s = −16t2 + 1054
= −16(4.5)² + 1054
= -16 * 20.25 + 1054
= 730
C. the time that it takes to strike the ground
s = −16t² + 1054
= 16t² = 1054
t² = 1054/16
= 65.88
t = √65.88
t = 8.11
Hence the time it would take to strike the ground is 8.11 seconds.
Read more on velocity here:
brainly.com/question/4931057
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