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To find the gradient of the function ( t = x^2y + e^z ) at the point ( P(1, 5, -2) ), we first need to compute the partial derivatives of ( t ) with respect to ( x ), ( y ), and ( z ).
1. Partial derivative with respect to ( x ):
[
frac{partial t}{partial x} = 2xy
]
2. Partial derivative with respect to ( y ):
[
frac{partial t}{partial y} = x^2
]
3. Partial derivative with respect to ( z ):
[
frac{partial t}{partial z} = e^z
]
Now, we will evaluate these partial derivatives at the point ( P(1, 5, -2) ).
– Calculate ( frac{partial t}{partial x} ) at ( P(1, 5, -2) ):
[
frac{partial t}{partial x}bigg|_{(1, 5, -2)} = 2(1)(5) = 10
]
– Calculate ( frac{partial t}{partial y} ) at ( P(1, 5, -2) ):
Answer: b
Explanation: Grad(t) = 2xy i + x2
j + ez k. On substituting p(1,5,-2), we get 10i + j +
0.135k.
Answer: b
Explanation: Grad(t) = 2xy i + x2 j + ez k. On substituting p(1,5,-2), we get 10i + j +
0.135k.