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MAT 275 – Spring 2009 Homework
Solution Paths |
| #28 | To avoid confusion between the x for a point on the graph and x in the equation of the tangent line it might be helpful to temporarily use x0 for the former. The slope of the tangent is g'(x0). The equation of the tangent line is thus y=g(x0)+g'(x0)(x-x0). Plug in (x0/2,0) to get the ODE. |
| #36 | P–N is the number of persons who do not have the disease so the ODE is dN/dt=N(P–N). |
| #46 | From Example 7 we know that the solutions to dy/dt=y2 are y(t)=1/(C–t). Running through the domain k times as fast for the same function will increase the rate by a factor of k. That is d/dt[y(kt)] = k dy/dt(kt) = k[y(kt)]2. So v(t)=1/(C–kt) are solutions to the ODE dv/dt=kv2. Use the initial condition v(0)=10 to get C. Use the fact that dv/dt=–1 when v=5 to determine k. Then solve v(t)=1 for t. |
| #39 | Substitute the given values and
compute the antiderivative. Use the initial condition y(–1/2)=0
to determine C. Then use the final condition y(1/2)=1
to determine vs. |
| W |
The force due to friction is F=μN
where N is the normal force.
Since N is the force of
gravity on the mass of the car, it will be constant as long as the
surface remains flat (or at a constant slope). The friction coefficient
μ is based on the
surface materials interacting and area of contact (tire and road), so
it will remain constant as long as none of those change. Thus we can
assume F is constant in these
problems. Using Newton's 2nd law, F=ma,
so as long as the mass of the car is constant, a will also be constant. |
| #64 | This uses Torricelli's law, however you do not need to solve the ODE. You are given that you want dy/dt to be constant. Plugging that constant into the ODE yields an simple algebraic equation. Plug in A(y)=πr2 and g=32 ft/s2 (converted to match your other units) to get an equation relating y and r and including the constant a. Use r(4)=1 to determine a. |
| #69 | Follow the instructions in the
problem to convert the second order ODE into a first order separable
ODE. The LHS will involve integrating 1/sqrt(1+v2). To do this, use the
hyperbolic trig substitution v=sinh(t). Since cosh2(t)=1+sinh2(t) and d/dt[sinh(t)]=cosh(t), the integral just becomes the
integral of 1. So the LHS becomes t
after integration. Taking sinh of both sides results in v. |