KAAP686 Mathematics for Biomechanics
Calculus review 1
Derivatives
Derivative of a function = slope of a function, how fast it is changing.
d(mt)/dt = m (make a picture)
d(mt + b) = m
d(t2)/dt = 2t (make a picture)
d(t2 + b) = 2t
d((t-t0)2 + b)/dt = …
d(t3)/dt = …
d(tn)/dt = ntn-1
d(f(t) + g(t))/dt = d(f(t))/dt + d(g(t))/dt
d(a f(t))/dt = a d(f(t))/dt [a = constant, independent of t]
The above two results are also the requirements for a function to be linear. Since differentiation meets these reqs, differentiation is a linear function.
Product, quotient
d(u(t) · v(t))/dt = [ u(t) · dv(t)/dt + v(t) · du(t)/dt ]
d(u(t)/v(t))/dt = [ v(t) · du(t)/dt – u(t) · dv(t)/dt ] / [v2(t)][v2(t)=v(t)*v(t)]
Derivatives of special functions
d(sin x)/dx =
d(cos x)/dx =
d(ex)/dx =
Chain Rule
Frequently (usually), equations involving special functions (such as sine, cosine, exponential) are more complicated than in examples above. For example you may have y(t)=Acos(ωt + φ) (where A, ω, and φ are constants). This is an example of a function that is “a function of a function”. The chain rule tells us how to differentiate such a function.
If y(t)=y(u), and u=g(t), then
dy/dt = (dy/du) · (du/dt)
At the end, since you don’t want u in the answer, substitute u(t) for u.
Example 1, chain rule:
y(t) = A cos(ωt + φ)[A, ω, and φ are constants]
What is dy/dt?
This can be thought of as y(u) = A*cos u and u(t) = ωt + φ.
Using chain rule:
dy/dt = dy(u)/du · du(t)/dt
= d (A cos u)/du · d(ωt + φ)/dt
= -A sin u · ω
= -Aω sin u
= -Aωsin(ωt + φ) {here I have replaced “u” with “u(t)” where u(t) = ωt + φ
(Others for HW.)
Example 2, principles of differentiation
Power = force * velocity (units??) = rate of doing (external) work
Power = F*v [muscle with force=F, shortening speed=v (positive=shortening)]
Tradeoff between F and v. Which tradeoff maximizes power?
A.V. Hill measured F, v experimentally and found that equation below gives a good fit to F, v data from concentric contractions, if a, b, and RHS constant are adjusted appropriately (Hill A.V. (1938) Proc Roy Soc B. 126: 136-195.)1
(F+a)*(v+b) = constant
Make graph. Discuss units for a and b; meaning of the intercepts of the curve with the F and v axes. Express RHS constant in terms of a, b, and axis intercepts.
(F+a)*(v+b) = b*(F0 +a)
Find the F, v combination which maximizes power (Fmp, vmp).
P = F*v = F(v)* v = …
When you get an expression for P(v), check the units.
How to use a derivative in this problem?
When you get expressions for Fmp and vmp, check the units.
If a = F0 /5 and b = vmax/5, what are Fmp and vmp, relative to F0 and vmax?
What is the maximum power, in terms of F0 and vmax?
1. The Hill equation is actually not that accurate. It makes a nice example due to its relative simplicity.
Example 3, principles of differentiation
This example is like Example 2 in that it involves the force-velocity relationship of muscle. It uses a more complicated equation that fits the experimental data better. (Marsh RL, Bennett AF (1986). J ExpBiol 126: 63-77.)
Marsh and Bennett (1986) collected data from the fast-glycolytic portion of the iliofibularis muscle of a small lizard. Numbers reported here are for experiments at 35°C.
Marsh & Bennett fitting their experimental force-velocity data to the Hill equation and to several more complicated equations. The equation (or model) that gave the best fit (minimum MSE) and most robust fit (parameters least sensitive to the removal of some of the experimental data) was the following:
V = B(l–F/F0)/ (A + F/F0) + C(1–F/F0)Eq. 1
Marsh & Bennett call this a Hill equation with a linear term added. The addition of the linear term (one additional parameter) improved the fit (i.e. reduced the MSE) by at least a factor of seven.
Measured Mean Values
F0 = maximum tetanic isometric force = 187 kN/m2.
Muscle cross sectional area = 1.2 mm2 = 1.2 x 10-6 m2 (2)
L0 = muscle length = 14.0 ± 0.3 mm = 1.40 x 10-2 m
Fitted Model Parameters
A / B (L0s-1) / C (L0s-1)0.113±0.010 / 1.64±0.20 / 7.52±0.36
where A is dimensionless and B and C have units of velocity, in “muscle lengths per second”.
Vmax and Wdotmaxwere computed for each muscle, using the fitted values of A, B, and C for that muscle. The mean values of Vmax and Wdotmax/(VmaxF0) are below.
Computed Model Parameters
Vmax (L0s-1) / Wdotmax/VmaxF021.9±0.6 / 0.118±0.002
where:
Vmax = predicted maximal shortening velocity
Wdot =dW/dt= power
Wdotmax =(dW/dt)max= max power
Wdotmax/(F0Vmax) = “curvature of the force-velocity relation”
Questions:
1. What is the predicted value for Vmax, the maximal velocity of shortening, in terms of A, B, and C in equation 1? (Hint: Remember that velocity of shortening is fastest when the force is zero. Use this knowledge to simplify equation 1 for the situation of maximum velocity.) Your answer should be a formula for Vmax that involves A, B, and/or C.
2. Use your answer to question 1 to compute the theoretical mean Vmax. Remember to include units. (Your answer should be close to the value given above, but might not be exactly equal.)
3. Make a plot (in Excel or another program) of predicted force (F)versus velocity (V). Use equation 1 to predict the force, with the mean values for F0, A, B, and C above. (You will not need to use the cross sectional area or the muscle length.) Units for F should be kN/m2; units for V should be L0/s. I suggest you compute the force at 5% increments from V=zero to V=Vmax.
4. Add a “power” column to your question 3 spreadsheet. Compute the power (P=F*V) at each velocity. Plot power versus velocity. (This plot will look like the plot for question 3, except it will have power instead of force on the vertical axis.) Note the maximum power. What is the velocity, Vmp,graph, at this “maximum power” point?
5. Find a formula for the value of velocity (V) that will give the maximum power. This formula will involve A, B, and/or C. Hints: First write the formula for power, P, as a function of V. Do this by multiplying the right hand side of equation 1 by V. Second, take the derivative, with respect to V, of the power formula. This will give you a formula for dP/dV that involves A, B, C, F0, and V. Now assume that dP/dV=0, i.e. put a 0 on the left hand side of the formula for dP/dV. Now solve for V, i.e. find the value of V that makes the right hand side equal zero. You can call your answer Vmp, for Vmaxpower, because this is the value of V that maximizes the power. Your answer for Vmp will be a formula involving A, B, C, and/or F0. If your formula for Vmp has a ± sign in it, you do not need to worry and you do not need to simplify it. It means there is more than one theoretical solution. However, it is likely that only one solution will be “reasonable”. This solution is the value of V that maximizes power, P, because this is the value of V that has dP/dV=0. Your formula for Vmp should be a formula such as Vmp = A*(B ± C) (but that is not the actual answer).
6. Use your answer to question 5 and the values for A, B, C above to compute a numeric value for Vmp. If your answer to question 5 has one or more ± signs in it, then compute Vmp using all possible combinations, and pick the answer than seems most reasonable. (For example, the answer for Vmp should not be negative, or a complex number, or infinity.) How big is Vmp compared to Vmax? How does Vmp compare to Vmp,graph? (It should be close.)
spreadsheet from 3
What are the Power = F*v [muscle with force=F, shortening speed=v (positive=shortening)]
Tradeoff between F and v. Which tradeoff maximizes power?
Make graph. Discuss units for a and b; meaning of the intercepts of the curve with the T and v axes. Express RHS constant in terms of a, b, and axis intercepts.
The maximum shorteneningvelcoity
Find the T, v combination which maximizes power (Tmp, vmp).
P = T*v = T(v)* v = …
When you get an expression for P(v), check the units.
How to use a derivative in this problem?
When you get expressions for Tmp and vmp, check the units.
If a = T0 /5 and b = vmax/5, what are Tmp and vmp, relative to T0 and vmax?
What is the maximum power, in terms of T0 and vmax?
2. Mean cross sectional area not specified. CSA was computed as mass/length. Mean mass = 16.8 ± 1.4 mg. Since the SDs are small, we can estimate the mean CSA as mean mass/mean length, assuming density=1 g/cm3.
Copyright © 2017 William C. Rose