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multi-variable min/max |
second partial derivative test |
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Single Variable Calculus |
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Multi-Variable Calculus |
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Acceleration Vector |
Arc Length (Vector Functions) |
Arc Length Function |
Arc Length Parameter |
Conservative Vector Fields |
Cross Product |
Curl |
Curvature |
Cylindrical Coordinates |
Lagrange Multipliers |
Line Integrals |
Partial Derivatives |
Partial Integrals |
Path Integrals |
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Principal Unit Normal Vector |
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This is the second exam for third semester (multi-variable) calculus. This exam contains 7 questions, one basic partial derivative, one chain rule, two directional derivatives, one lagrange multiplier, one multi-variable min/max and one tangent plane.
Each exam page contains a full exam with detailed solutions. Most of these are actual exams from previous semesters used in college courses. You may use these as practice problems or as practice exams. Here are some suggestions on how to use these to help you prepare for your exams.
- Set aside a chunk of full, uninterrupted time, usually an hour to two, to work each exam.
- Go to a quiet place where you will not be interrupted that duplicates your exam situation as closely as possible.
- Use the same materials that you are allowed in your exam (unless the instructions with these exams are more strict).
- Use your calculator as little as possible except for graphing and checking your calculations.
- Work the entire exam before checking any solutions.
- After checking your work, rework any problems you missed and go to the 17calculus page discussing the material to perfect your skills.
- Work as many practice exams as you have time for. This will give you practice in important techniques, experience in different types of exam problems that you may see on your own exam and help you understand the material better by showing you what you need to study.
IMPORTANT -
Exams can cover only so much material. Instructors will sometimes change exams from one semester to the next to adapt an exam to each class depending on how the class performs during the semester while they are learning the material. So just because you do well (or not) on these practice exams, does not necessarily mean you will do the same on your exam. Your instructor may ask completely different questions from these. That is why working lots of practice problems will prepare you better than working just one or two practice exams.
Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.
Exam Details | |
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Time | 2 hours |
Questions | 7 |
Total Points | 75 |
Tools | |
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Calculator | no |
Formula Sheet(s) | 1 page, 8.5x11 or A4 |
Other Tools | none |
(5 points) Calculate the first partial derivatives of \( f(u,v,t) = e^{uv} \sin(ut) \).
Problem Statement |
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(5 points) Calculate the first partial derivatives of \( f(u,v,t) = e^{uv} \sin(ut) \).
Solution |
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\(\displaystyle{\frac{\partial f}{\partial u}} = te^{uv}\cos(ut) + ve^{uv}\sin(ut) \) |
\(\displaystyle{\frac{\partial f}{\partial v}} = ue^{uv}\sin(ut) \) |
\(\displaystyle{\frac{\partial f}{\partial t}} = ue^{uv}\cos(ut) \) |
Not much explanation is required here. The first partial derivative requires the product rule. The other two are basic partial derivatives.
close solution |
(10 points) Use Lagrange multipliers to find the point(s) on the surface \(z^2-x^2-y^2=0\) that are closest to the point \((1,7,0)\).
Problem Statement |
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(10 points) Use Lagrange multipliers to find the point(s) on the surface \(z^2-x^2-y^2=0\) that are closest to the point \((1,7,0)\).
Final Answer |
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The two points \((1/2, 7/2, \pm 5/\sqrt{2})\) on the surface \(z^2-x^2-y^2=0\) are closest to the point \((1,7,0)\). |
Problem Statement |
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(10 points) Use Lagrange multipliers to find the point(s) on the surface \(z^2-x^2-y^2=0\) that are closest to the point \((1,7,0)\).
Solution |
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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.
\(g(x,y,z) = z^2-x^2-y^2=0\) 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 \(d=\sqrt{(x-1)^2+(y-7)^2+(z-0)^2}\). This is the function to be minimized. To simplify the function, we can use the square of this and minimize \(f(x,y,z)=(x-1)^2+(y-7)^2+z^2\)
Now we calculate \( \nabla f = \lambda \nabla g \).
\( \nabla f = 2(x-1)\hat{i} + 2(y-7)\hat{j} + 2z\hat{k} \)
\( \nabla g = -2x\hat{i} -2y\hat{j} +2z\hat{k} \)
Using the equation \( \nabla f = \lambda \nabla g \) and equating the i, j, k components, we get
\(2(x-1) = -2\lambda x\) |
\(2(y-7) = -2\lambda y\) |
\(2z = 2\lambda z\) |
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 \(g(x,y,z) = z^2-x^2-y^2=0\) and solve for \(z=\pm 5/\sqrt{2}\).
Final Answer |
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The two points \((1/2, 7/2, \pm 5/\sqrt{2})\) on the surface \(z^2-x^2-y^2=0\) are closest to the point \((1,7,0)\). |
close solution |
(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 \((2,1,4)\).
Problem Statement |
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(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 \((2,1,4)\).
Solution |
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Let \(F(x,y,z)=x^2y-z\).
\(\displaystyle{ \nabla F(x,y,z) = \frac{\partial F}{\partial x}\hat{i} + \frac{\partial F}{\partial y}\hat{j} + \frac{\partial F}{\partial z}\hat{k} = }\)
\(2xy\hat{i}+x^2\hat{j}-\hat{k}\)
\(\nabla F(2,1,4) = 4\hat{i}+4\hat{j}-\hat{k}\)
This vector \( \nabla F(2,1,4) \) is normal to the tangent plane. So the equation of the plane can be found using
\( \langle 4,4,-1 \rangle \cdot \langle x-2, y-1, z-4 \rangle = 0 \) |
\( 4(x-2)+4(y-1)-(z-4) = 0 \) |
\( 4x-8 + 4y - 4 -z + 4 = 0 \) |
\(\boxed{ 4x+4y-z=8 }\)
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
\(\boxed{x=2+4t ~~~~~ y=1+4t ~~~~~ z=4-t}\)
close solution |
(10 points) If \(u=x^2-2y^2+z^3\) 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).
Problem Statement |
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(10 points) If \(u=x^2-2y^2+z^3\) 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).
Final Answer |
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\(\displaystyle{ \frac{du}{dt} = 2\sin(t)\cos(t) - 4e^{2t} + 81t^2 }\) |
Problem Statement |
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(10 points) If \(u=x^2-2y^2+z^3\) 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).
Solution |
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\(\displaystyle{\frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt} + \frac{\partial u}{\partial z} \frac{dz}{dt}}\)
\(\displaystyle{\frac{\partial u}{\partial x} = 2x}\) |
\(\displaystyle{\frac{dx}{dt} = \cos(t)}\) | |
\(\displaystyle{\frac{\partial u}{\partial y} = -4y}\) |
\(\displaystyle{ \frac{dy}{dt} = e^t}\) | |
\(\displaystyle{\frac{\partial u}{\partial z} = 3z^2}\) |
\(\displaystyle{\frac{dz}{dt} = 3}\) |
\(\displaystyle{ \frac{du}{dt} } = 2x\cos(t) + (-4y)e^t +3z^2 (3) = 2\sin(t)\cos(t) -4e^t(e^t) + 3(3t)^2(3) \)
Final Answer |
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\(\displaystyle{ \frac{du}{dt} = 2\sin(t)\cos(t) - 4e^{2t} + 81t^2 }\) |
close solution |
(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.
Problem Statement |
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(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.
Solution |
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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
\(\vec{v} = (0+1)\hat{i} + (3-2)\hat{j} + (3-1)\hat{k} = \hat{i}+\hat{j}+2\hat{k}\)
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.
close solution |
(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 \((2,1)\).
Problem Statement |
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(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 \((2,1)\).
Final Answer |
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\(\displaystyle{\frac{\partial z}{\partial x} = \frac{9}{5} ~~~~~ \frac{\partial z}{\partial y} = \frac{-1}{10}}\) |
Problem Statement |
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(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 \((2,1)\).
Solution |
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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 \((2,1) \to (1,3)\).
\(\displaystyle{\hat{u} = \frac{(1-2)\hat{i}+(3-1)\hat{j}}{\sqrt{1+2^2}} = \frac{-\hat{i}+2\hat{j}}{\sqrt{5}}}\)
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.
\(\displaystyle{\hat{v} = \frac{\langle 5-2, 5-1 \rangle}{\sqrt{3^2+4^2}} = \frac{\langle 3,4 \rangle}{5}}\)
\(\displaystyle{\langle a,b \rangle \cdot \frac{\langle 3,4 \rangle}{5} = 1}\)
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
\(\displaystyle{a = \frac{9}{5} ~~~ b = \frac{-1}{10}}\)
Final Answer |
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\(\displaystyle{\frac{\partial z}{\partial x} = \frac{9}{5} ~~~~~ \frac{\partial z}{\partial y} = \frac{-1}{10}}\) |
close solution |
(10 points) Find the point(s) on the surface \(z^2-x^2-y^2=0\) 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.
Problem Statement |
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(10 points) Find the point(s) on the surface \(z^2-x^2-y^2=0\) 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.
Final Answer |
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The two points \((1/2, 7/2, \pm 5/\sqrt{2})\) on the surface \(z^2-x^2-y^2=0\) are closest to the point \((1,7,0)\). |
Problem Statement |
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(10 points) Find the point(s) on the surface \(z^2-x^2-y^2=0\) 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.
Solution |
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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 \(d=\sqrt{(x-1)^2+(y-7)^2+(z-0)^2}\). 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 \(r=(x-1)^2+(y-7)^2+z^2\). 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 \(z^2-x^2-y^2=0\) 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 \(r=(x-1)^2+(y-7)^2+x^2+y^2\)
\(r_x=2(x-1)+2x\)
\(r_x=0 \to 2(x-1)+2x=0 \to x=1/2\)
\(r_y=2(y-7)+2y\)
\(r_y=0 \to 2(y-7)+2y=0 \to y=7/2\)
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.
Final Answer |
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The two points \((1/2, 7/2, \pm 5/\sqrt{2})\) on the surface \(z^2-x^2-y^2=0\) are closest to the point \((1,7,0)\). |
close solution |