You CAN Ace Calculus
external links you may find helpful |
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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 |
Potential Functions |
Principal Unit Normal Vector |
Differential Equations |
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Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
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For the following discussion, we will consider a parameterized curve defined by the vector function \(\displaystyle{ \vec{r}(t) = \langle x(t), y(t), z(t) \rangle }\) which is traversed once on the continuous interval \( a \leq t \leq b \).
Some books use the Greek letter \(\kappa\) (kappa) for curvature. We use a capital K.
The curvature of a smooth curve is a measure of how 'tight' or 'sharp' the curve is. If we have a smooth curve \(\vec{r}\) and we have a function s which is the arc length function, the curvature is defined to be
\(\displaystyle{ K(s) = \left\| \frac{d\vec{T}}{ds} \right\|}\).
This equation for the curvature is not particularly useful for calculations. So we have several other ways to write the equation of the curvature. But first notice, that the curvature is a scalar function, not a vector function. And since it is the norm of a vector, the curvature will always be positive.
Curvature Formula #1 |
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For our first equation to use when calculating the curvature, we will use the chain rule to write \(\displaystyle{ \frac{d\vec{T}}{dt} = \frac{d\vec{T}}{ds} \cdot \frac{ds}{dt} }\). We can solve for \(d\vec{T}/ds\) to get
\(\displaystyle{ K = \left\| \frac{d\vec{T}}{ds} \right\| = \frac{\| d\vec{T}/dt \|}{ \| ds/dt \| } = \frac{1}{\| \vec{v} \|} \left\| \frac{d\vec{T}}{dt} \right\| }\)
Notice in the previous equation, we used \(ds/dt = \|\vec{v}\|\) to simplify the equation somewhat.
Now we can write the first curvature formula in a form that we can use for calculations.
\(\displaystyle{ K(t) = \frac{1}{\|\vec{v}\|} \left\| \frac{d\vec{T}}{dt} \right\| = \frac{\| \vec{T}'(t) \|}{\|\vec{r}'(t)\|} }\)
Curvature Formula #2 |
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If we define a vector \(\vec{a} = d\vec{v}/dt\) as an acceleration vector, a second curvature formula is
\(\displaystyle{ K=\frac{\|\vec{v} \times \vec{a} \|}{\|\vec{v}\|^3} }\)
Before working some practice problems, here is a quick video clip for you that should help you understand curvature a bit better.
video by MIP4U
Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems |
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Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on. |
Instructions - - Unless otherwise instructed, calculate the curvature of these vector functions. Give your answers in exact, completely factored form.
\( \vec{r}(t) = 3t\vhat{i} + 4\sin t\vhat{j} + \) \( 4\cos t\vhat{k} \)
Problem Statement |
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\( \vec{r}(t) = 3t\vhat{i} + 4\sin t\vhat{j} + \) \( 4\cos t\vhat{k} \)
Final Answer |
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\( K(t) = 4/25 \) |
Problem Statement |
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\( \vec{r}(t) = 3t\vhat{i} + 4\sin t\vhat{j} + \) \( 4\cos t\vhat{k} \)
Solution |
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video by Krista King Math
Final Answer |
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\( K(t) = 4/25 \) |
close solution |
\( \vec{r}(t) = \langle 2\cos(t), 2\sin(t) \rangle \)
Problem Statement |
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\( \vec{r}(t) = \langle 2\cos(t), 2\sin(t) \rangle \)
Final Answer |
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\( K(t) = 1/2 \) |
Problem Statement |
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\( \vec{r}(t) = \langle 2\cos(t), 2\sin(t) \rangle \)
Solution |
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video by MIP4U
Final Answer |
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\( K(t) = 1/2 \) |
close solution |
Find the curvature at \(t=1\) for \( \vec{r}(t) = \langle t, t^2/2, t^3/3 \rangle \).
Problem Statement |
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Find the curvature at \(t=1\) for \( \vec{r}(t) = \langle t, t^2/2, t^3/3 \rangle \).
Final Answer |
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\( K(1) = \sqrt{2}/3 \) |
Problem Statement |
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Find the curvature at \(t=1\) for \( \vec{r}(t) = \langle t, t^2/2, t^3/3 \rangle \).
Solution |
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video by MIP4U
Final Answer |
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\( K(1) = \sqrt{2}/3 \) |
close solution |
Find the curvature of \( \vec{r}(t) = \langle -5t, 2t^3, 3t^4 \rangle \) at \( t = 2 \).
Problem Statement |
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Find the curvature of \( \vec{r}(t) = \langle -5t, 2t^3, 3t^4 \rangle \) at \( t = 2 \).
Final Answer |
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\( K(t) = 24\sqrt{3229}/(9817)^{3/2} \) |
Problem Statement |
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Find the curvature of \( \vec{r}(t) = \langle -5t, 2t^3, 3t^4 \rangle \) at \( t = 2 \).
Solution |
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video by MIP4U
Final Answer |
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\( K(t) = 24\sqrt{3229}/(9817)^{3/2} \) |
close solution |
Find the curvature of \( \vec{r}(t) = \langle 2\cos(3t),2\sin(3t), \) \(4t \rangle \) at \( t = 0 \).
Problem Statement |
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Find the curvature of \( \vec{r}(t) = \langle 2\cos(3t),2\sin(3t), \) \(4t \rangle \) at \( t = 0 \).
Final Answer |
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9/26 |
Problem Statement |
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Find the curvature of \( \vec{r}(t) = \langle 2\cos(3t),2\sin(3t), \) \(4t \rangle \) at \( t = 0 \).
Solution |
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This problem is solved in each of these videos. The first video uses the cross product version of the curvature formula, the second video does not use the cross product.
video by MIP4U
video by MIP4U
Final Answer |
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9/26 |
close solution |