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Given that Z = x2 + y2 + x2y', Evaluate (i) dz/dx (ii) dz/dy, when x = 1 and y = 1. If y = ae^x + be2x, show that y'' - 3y' + 2y = 0. A body moves in a straight line according to the law of motion s = t3 − 2t2 + t. Find velocity and acceleration after 3 seconds.

Mathematics

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here's the solution for Question 5a:

(5a). Given that Z=x2+y2+x2y3Z = x^2 + y^2 + x^2y^3, Evaluate (i) ∂Z∂x\frac{\partial Z}{\partial x} (ii) ∂Z∂y\frac{\partial Z}{\partial y}, when x=1x=1 and y=1y=1

  1. Find the partial derivative of ZZ with respect to xx, denoted as ∂Z∂x\frac{\partial Z}{\partial x}.

    When differentiating with respect to xx, treat yy as a constant.

    Z=x2+y2+x2y3Z = x^2 + y^2 + x^2y^3 ∂Z∂x=∂∂x(x2)+∂∂x(y2)+∂∂x(x2y3)\frac{\partial Z}{\partial x} = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial x}(y^2) + \frac{\partial}{\partial x}(x^2y^3) ∂Z∂x=2x+0+2xy3\frac{\partial Z}{\partial x} = 2x + 0 + 2xy^3 ∂Z∂x=2x+2xy3\frac{\partial Z}{\partial x} = 2x + 2xy^3
  2. Evaluate ∂Z∂x\frac{\partial Z}{\partial x} at x=1x=1 and y=1y=1.

    Substitute x=1x=1 and y=1y=1 into the expression for ∂Z∂x\frac{\partial Z}{\partial x}.

    ∂Z∂x∣(1,1)=2(1)+2(1)(1)3\frac{\partial Z}{\partial x} \Big|_{(1,1)} = 2(1) + 2(1)(1)^3 ∂Z∂x∣(1,1)=2+2(1)(1)\frac{\partial Z}{\partial x} \Big|_{(1,1)} = 2 + 2(1)(1) ∂Z∂x∣(1,1)=2+2\frac{\partial Z}{\partial x} \Big|_{(1,1)} = 2 + 2 ∂Z∂x∣(1,1)=4\frac{\partial Z}{\partial x} \Big|_{(1,1)} = \boxed{4}
  3. Find the partial derivative of ZZ with respect to yy, denoted as ∂Z∂y\frac{\partial Z}{\partial y}.

    When differentiating with respect to yy, treat xx as a constant.

    Z=x2+y2+x2y3Z = x^2 + y^2 + x^2y^3 ∂Z∂y=∂∂y(x2)+∂∂y(y2)+∂∂y(x2y3)\frac{\partial Z}{\partial y} = \frac{\partial}{\partial y}(x^2) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial y}(x^2y^3) ∂Z∂y=0+2y+x2(3y2)\frac{\partial Z}{\partial y} = 0 + 2y + x^2(3y^2) ∂Z∂y=2y+3x2y2\frac{\partial Z}{\partial y} = 2y + 3x^2y^2
  4. Evaluate ∂Z∂y\frac{\partial Z}{\partial y} at x=1x=1 and y=1y=1.

    Substitute x=1x=1 and y=1y=1 into the expression for ∂Z∂y\frac{\partial Z}{\partial y}.

    ∂Z∂y∣(1,1)=2(1)+3(1)2(1)2\frac{\partial Z}{\partial y} \Big|_{(1,1)} = 2(1) + 3(1)^2(1)^2 ∂Z∂y∣(1,1)=2+3(1)(1)\frac{\partial Z}{\partial y} \Big|_{(1,1)} = 2 + 3(1)(1) ∂Z∂y∣(1,1)=2+3\frac{\partial Z}{\partial y} \Big|_{(1,1)} = 2 + 3 ∂Z∂y∣(1,1)=5\frac{\partial Z}{\partial y} \Big|_{(1,1)} = \boxed{5}
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