![f(x,y,z)=2z^2x+y^3](https://tex.z-dn.net/?f=f%28x%2Cy%2Cz%29%3D2z%5E2x%2By%5E3)
has gradient
![\nabla f(x,y,z)=2z^2\,\vec\imath+3y^2\,\vec\jmath+4xz\,\vec k](https://tex.z-dn.net/?f=%5Cnabla%20f%28x%2Cy%2Cz%29%3D2z%5E2%5C%2C%5Cvec%5Cimath%2B3y%5E2%5C%2C%5Cvec%5Cjmath%2B4xz%5C%2C%5Cvec%20k)
which at the point (-1, 4, 3) has a value of
![\nabla f(-1,4,3)=18\,\vec\imath+48\,\vec\jmath-12\,\vec k](https://tex.z-dn.net/?f=%5Cnabla%20f%28-1%2C4%2C3%29%3D18%5C%2C%5Cvec%5Cimath%2B48%5C%2C%5Cvec%5Cjmath-12%5C%2C%5Cvec%20k)
I'm not sure what the given direction vector is supposed to be, but my best guess is that it's intended to say
, in which case we have
![\|\vec u\|=\sqrt{15^2+25^2}=5\sqrt{34}](https://tex.z-dn.net/?f=%5C%7C%5Cvec%20u%5C%7C%3D%5Csqrt%7B15%5E2%2B25%5E2%7D%3D5%5Csqrt%7B34%7D)
Then the derivative of
at (-1, 4, 3) in the direction of
is
![D_{\vec u}f(-1,4,3)=\nabla f(-1,4,3)\cdot\dfrac{\vec u}{\|\vec u\|}=\boxed{\dfrac{294}{\sqrt{34}}}](https://tex.z-dn.net/?f=D_%7B%5Cvec%20u%7Df%28-1%2C4%2C3%29%3D%5Cnabla%20f%28-1%2C4%2C3%29%5Ccdot%5Cdfrac%7B%5Cvec%20u%7D%7B%5C%7C%5Cvec%20u%5C%7C%7D%3D%5Cboxed%7B%5Cdfrac%7B294%7D%7B%5Csqrt%7B34%7D%7D%7D)
Answer:
225 square centimeters
Step-by-step explanation:
Use the distance formula?
1.12=1 12/100=1 3/25
or 28/25