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Calculus 3  Exam 2 ( Semester A ) 
Exam Overview 
This is the second exam for third semester (multivariable) calculus. This exam contains 7 questions, one basic partial derivative, one chain rule, two directional derivatives, one lagrange multiplier, one multivariable min/max and one tangent plane. 

Exam Details 

Tools 
Time  2 hours  
Calculators  no 
Questions  7  
Formula Sheet(s) 
1 page, 8.5x11 or A4 
Total Points  75  
Other Tools  none 
Instructions:
 Show all your work.
 For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.
 Correct notation counts (i.e. points will be taken off for incorrect notation).
 Give exact, simplified answers.
[ Click on the question to reveal/hide the solution. ]
Question 1 
(5 points) Calculate the first partial derivatives of \(f(u,v,t)=e^{uv}\sin(ut)\). 

\(\displaystyle{
\begin{array}{rcl}
\frac{\partial f}{\partial u} & = & te^{uv}\cos(ut)+ve^{uv}\sin(ut) \\
\frac{\partial f}{\partial v} & = & ue^{uv}\sin(ut) \\
\frac{\partial f}{\partial t} & = & ue^{uv}\cos(ut)
\end{array}
}\)
Not much explanation is required here. The first partial derivative requires the product rule. The other two are just basic partial derivatives.
Question 2 
(10 points) If \(u=x^22y^2+z^3\) and \(x=\sin(t), y=e^t, z=3t\), calculate \(du/dt\) using the appropriate chain rule. Make sure your final answer has only numbers and t (no x, y or z). 


\(\displaystyle{ \frac{du}{dt} = 2\sin(t)\cos(t)  4e^{2t} +81t^2 }\) 
\(\displaystyle{ \frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt} + \frac{\partial u}{\partial z} \frac{dz}{dt} }\)
\(\displaystyle{ \frac{\partial u}{\partial x} = 2x }\)  
\(\displaystyle{ \frac{dx}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{\partial u}{\partial y} = 4y }\)  
\(\displaystyle{ \frac{dy}{dt} = e^t }\) 
\(\displaystyle{ \frac{\partial u}{\partial z} = 3z^2 }\)  
\(\displaystyle{ \frac{dz}{dt} = 3 }\) 
\(\displaystyle{ \frac{du}{dt} = 2x\cos(t) + (4y)e^t +3z^2 (3) = }\)
\( 2\sin(t)\cos(t) 4e^t(e^t)+3(3t)^2(3) \)
Question 2 Final Answer 
\(\displaystyle{ \frac{du}{dt} = 2\sin(t)\cos(t)  4e^{2t} +81t^2 }\) 
Question 3 
(10 points) Calculate the directional derivative of \(f(x,y,z)=x^3+y^3z\) at \((1,2,1)\) in the direction toward the point \((0,3,3)\). Is this the maximum directional derivative? Explain and show work why or why not. 

We are given the function and the point but not the vector. However, we are given enough information to find the vector from the point \((1,2,1)\) in the direction toward the point \((0,3,3)\). So the vector is
\( \vec{v} = (0+1)\hat{i} + (32)\hat{j} + (31)\hat{k} = \hat{i}+\hat{j}+2\hat{k} \)
The directional derivative requires a unit vector.
The length of this vector is \( \\vec{v}\ = \sqrt{1+1+4} = \sqrt{6} \). To get a unit vector, we divide \(\vec{v}\) by the length.
\(\displaystyle{ \hat{u} = \frac{\vec{v}}{\\vec{v}\} = \frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{6}} }\)
The directional derivative of \(f(x,y,z)=x^3+y^3z\) in the direction \(\hat{u}\) is
\(D_{\hat{u}}f(x,y,z)=\nabla f \cdot \hat{u} = \)
\(\displaystyle{ \left( \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j} + \frac{\partial f}{\partial z} \hat{k} \right)
\cdot \hat{u} = }\)
\(\displaystyle{ \left( 3x^2 \hat{i} + 3y^2 z\hat{j} +y^3\hat{k} \right) \cdot \hat{u} }\)
Now, we need to find the directional derivative at the point \((1,2,1)\).
\(D_{\hat{u}}f(1,2,1)=\nabla f \cdot \hat{u} = \)
\(\displaystyle{ \left( 3\hat{i} +12\hat{j} +8\hat{k} \right) \cdot \frac{ \hat{i} +\hat{j} +2\hat{k} }{\sqrt{6}} = }\)
\(\displaystyle{ \frac{1}{\sqrt{6}}(3+12+16) = \frac{31}{\sqrt{6}} }\)
\(\boxed{\displaystyle{ D_{\hat{u}}f(1,2,1)= \frac{31}{\sqrt{6}} }}\)
The maximum directional derivative is the magnitude of the gradient.
From the work above, the gradient is
\( \nabla f(1,2,1) = 3\hat{i} +12\hat{j} + 8\hat{k} \)
\( \ \nabla f(1,2,1) \ = \sqrt{3^2+12^2+8^2} = \sqrt{217} \)
Since \(\displaystyle{ \frac{31}{\sqrt{6}} \neq \sqrt{217} }\), the directional derivative in the direction of \(\hat{u}\) is not the maximum directional derivative.
Question 4 
(15 points) The directional derivative of \(z=f(x,y)\) at \((2,1)\) in the direction toward the point \((1,3)\) is \(2/\sqrt{5}\) and the directional derivative in the direction toward the point \((5,5)\) is 1. Compute \(\partial z / \partial x\) and \(\partial z / \partial y\) at \((2,1)\). 


\(\displaystyle{ \frac{\partial z}{\partial x} = \frac{9}{5} ~~~~~ \frac{\partial z}{\partial y} = \frac{1}{10} }\) 
The directional derivative is the magnitude of the gradient in the direction of a unit vector. This can be written
\(\displaystyle{ \nabla f \cdot \hat{u} = \left( \frac{\partial f}{\partial x}\hat{i} +\frac{\partial f}{\partial y}\hat{j} \right) \cdot \hat{u} }\) and we are asked to find \(\partial z / \partial x\) and \(\partial z / \partial y\). To simplify the equations, we will set \( a=\partial z / \partial x\) and \( b= \partial z / \partial y\).
For the first direction, we have \((2,1) \to (1,3)\) and the directional derivative is \(2/\sqrt{5}\). So first we need to find the unit vector from \((2,1) \to (1,3)\).
\(\displaystyle{ \hat{u} = \frac{(12)\hat{i}+(31)\hat{j}}{\sqrt{1+2^2}} = \frac{\hat{i}+2\hat{j}}{\sqrt{5}} }\)
So now we can put together the directional derivative equation (for simplicity, we will switch to the bracket notation)
\(\displaystyle{ \langle a,b \rangle \cdot \frac{\langle 1,2 \rangle}{\sqrt{5}} = \frac{2}{\sqrt{5}} }\)
Now we do the same steps for the other directional derivative.
\(\displaystyle{ \hat{v} = \frac{\langle 52, 51 \rangle}{\sqrt{3^2+4^2}} = \frac{\langle 3,4 \rangle}{5}}\)
\(\displaystyle{ \langle a,b \rangle \cdot \frac{\langle 3,4 \rangle}{5} = 1 }\)
From the two dot products, we get two equations with two unknowns, as follows.
\(\displaystyle{ \frac{a}{\sqrt{5}} + \frac{2b}{\sqrt{5}} = \frac{2}{\sqrt{5}} }\) 

\(\displaystyle{ \frac{3a}{5} + \frac{4b}{5} =1 }\) 
\(a+2b=2\) 

\(3a+4b=5\) 
Solving for a and b, we get
\(\displaystyle{ a = \frac{9}{5} ~~~ b = \frac{1}{10} }\)
Question 4 Final Answer 
\(\displaystyle{ \frac{\partial z}{\partial x} = \frac{9}{5} ~~~~~ \frac{\partial z}{\partial y} = \frac{1}{10} }\) 
Question 5 
(10 points) Use Lagrange multipliers to find the point(s) on the surface \(z^2x^2y^2=0\) that are closest to the point \((1,7,0)\). 


The two points \((1/2, 7/2, \pm 5/\sqrt{2})\) on the surface \(z^2x^2y^2=0\) are closest to the point \((1,7,0)\). 
For Lagrange multipliers we use the equations \( \nabla f = \lambda \nabla g \) and \(g=0\) where \(g(x,y,z)\) is the constraint equation and \(f(x,y,z)\) is the equation to be minimized or maximized.
\(g(x,y,z) = z^2x^2y^2=0\) is given in the problem statement.
If we let \((x,y,z)\) represent any point on the function \(f(x,y,z)\), then the distance between \((x,y,z)\) and \((1,7,0)\) is
\(d=\sqrt{(x1)^2+(y7)^2+(z0)^2}\). This is the function to be minimized. To simplify the function, we can use the square of this and minimize \(f(x,y,z)=(x1)^2+(y7)^2+z^2\)
Now we calculate \( \nabla f = \lambda \nabla g \).
\( \nabla f = 2(x1)\hat{i} + 2(y7)\hat{j} + 2z\hat{k} \)
\( \nabla g = 2x\hat{i} 2y\hat{j} +2z\hat{k} \)
Using the equation \( \nabla f = \lambda \nabla g \) and equating the i, j, k components, we get
\(2(x1) = 2\lambda x\) 
\(2(y7) = 2\lambda y\) 
\(2z = 2\lambda z\) 
The last equation can be solved for \(\lambda = 1\). Substituting this back into the other two equations, gives us \(x=1/2\) and \(y=7/2\). Finally, we use \(g(x,y,z) = z^2x^2y^2=0\) and solve for \(z=\pm 5/\sqrt{2}\).
Question 5 Final Answer 
The two points \((1/2, 7/2, \pm 5/\sqrt{2})\) on the surface \(z^2x^2y^2=0\) are closest to the point \((1,7,0)\). 
Question 6 
(10 points) Find the point(s) on the surface \(z^2x^2y^2=0\) that are closest to the point \((1,7,0)\) by calculating the critical points and using the Second Derivative Partials Test to conclude that your answer(s) are minimums. 


The two points \((1/2, 7/2, \pm 5/\sqrt{2})\) on the surface \(z^2x^2y^2=0\) are closest to the point \((1,7,0)\). 
This is the same question as the previous one but we need to solve it differently. We can also use the answers from the previous question to verify our work.
The function we need to minimize is the distance between a point \((x,y,z)\) and the point \((1,7,0)\). This equation is \(d=\sqrt{(x1)^2+(y7)^2+(z0)^2}\). As we did in the previous question, we can minimize \(r=d^2\) to make the algebra easier. So the equation we need to minimize if \(r=(x1)^2+(y7)^2+z^2\). To do so, we will calculate \(r_x\) and \(r_y\).
But first, we need to use that fact that the point \((x,y,z)\) is on the surface \(z^2x^2y^2=0\) to solve for \(z^2\) and substitute it into the first equation to remove the \(z\). This will give us an equation in only \(x\) and \(y\),
\(z^2=x^2+y^2\) and so \(r=(x1)^2+(y7)^2+x^2+y^2\)
\(r_x=2(x1)+2x\)
\(r_x=0 \to 2(x1)+2x=0 \to x=1/2\)
\(r_y=2(y7)+2y\)
\(r_y=0 \to 2(y7)+2y=0 \to y=7/2\)
Since \(z^2=x^2+y^2 \to z^2 = (1/2)^2+(7/2)^2 \to z=\pm 5/\sqrt{2}\)
So the two points are \((1/2, 7/2, \pm 5/\sqrt{2})\) which matches our answer from the previous question.
Now we need to use the second derivative partials test to verify these are minimum values.
\(r_{xx}=2+2=4 ~~~~~ r_{yy}=2+2=4 ~~~~~ r_{xy}=0\)
\(D(x,y) = r_{xx}r_{yy}[r_{xy}]^2 = 4(4)0=16\)
This value holds for both points, so since \(D(x,y)>0\) and \(r_{xx}=4>0\), the points are minimums.
Question 6 Final Answer 
The two points \((1/2, 7/2, \pm 5/\sqrt{2})\) on the surface \(z^2x^2y^2=0\) are closest to the point \((1,7,0)\). 
Question 7 
(10 points) Find the equation of the tangent plane and the parametric equations of the normal line to the surface \(f(x,y)=x^2y\) at the point \((2,1,4)\). 

Let \(F(x,y,z)=x^2yz\).
\(\displaystyle{ \nabla F(x,y,z) = \frac{\partial F}{\partial x}\hat{i} + \frac{\partial F}{\partial y}\hat{j} + \frac{\partial F}{\partial z}\hat{k} = }\)
\(2xy\hat{i}+x^2\hat{j}\hat{k}\)
\( \nabla F(2,1,4) = 4\hat{i}+4\hat{j}\hat{k} \)
This vector \( \nabla F(2,1,4) \) is normal to the tangent plane. So the equation of the plane can be found using
\(
\begin{array}{rcl}
\langle 4,4,1 \rangle \cdot \langle x2, y1, z4 \rangle & = & 0 \\
4(x2)+4(y1)(z4) & = & 0 \\
4x8 + 4y  4 z +4 & = & 0
\end{array}
\)
\(\boxed{ 4x+4yz=8 }\)
The parametric equations of the normal line are found from the normal vector \(4\hat{i}+4\hat{j}\hat{k} \) and the point \((2,1,4)\) as
\(\boxed{ x=2+4t ~~~~~ y=1+4t ~~~~~ z=4t }\)