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3s=2t can also be written as 3y=2x or 3x=2y. Either way, it is linear.
To find out if it is linear, simply graph it. If you can draw a completely vertical line through any point of the graph without intersecting more than one point of the graph, then it is linear.
This equation (3s=2t), it is linear.
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What is the integral of 2 times 1-x dx with limits 0 to t?
,/` 2(1 - x) dx,/` 2 - 2x dx2x - x2 ...evaluated from 0 to t gives us...2t - t2 - [2(0) - (0)2]2t - t2
What is the limit as t approaches 0 of 2t tan t?
That would be 0. (But that was too easy. Did you mean something else?)
How do you make a the subject of t equals 0.5 x a plus b x h?
To make the following relationship: t = (1/2)a +bh a function of "a" (a = ...) start by subtracting both sides by "bh": t - bh = (1/2)a then divide both sides by (1/2), which is the same as multiplying by 2: 2(t - bh) = a from here you can distribute the 2 if desired: a = 2t - 2bh
Integration of tangent cubed of x?
Find I = ∫ tan³ x dx. The solution is: I = ½ tan² x - log cos x. * * * Here is how we can obtain this result: First, let t = tan x, s = sin x, and c = cos x; then, dI = t³ dx, ds = c dx, dc = -s dx, and dt = (1 + t²) dx; and, of course, t = s / c. By algebra, t³ = t(t² + 1) - t; thus, we have dI = t³ dx = t(t² + 1) dx - t dx = t dt - t dx. Now, d (t²) = 2t dt; thus, t dt = ½ d(t²). On the other hand, we have d log c = dc / c = -s dx / c = -t dx; thus, t dx = -d log c. Combining these results, we have dI = t dt - t dx = ½ d(t²) - d log c. This integrates readily, giving I = ½ t² - log c, which is the solution we sought. * * * We may check our result, by differentiating back: dt / dx = 1 + t²; and d(t²) / dt = 2t; thus, (d/dx)(t²) = 2t dt / dx = 2t (1 + t²). Also, we have d log c / dc = 1 / c; and dc / dx = -s; whence, (d/dx)(log c) = (dc / dx) / c = -s / c = -t. Then, dI / dx = ½ (d/dx)(t²) - (d/dx)(log c) = t (1 + t²) - t = t + t³ - t = t³, re-assuring us that we have integrated correctly.
A hemi-spherical vessel has internal radius 0.5 m It is initially empty Water flows in at a constant rate of 1 liter per second Find an expression for the depth of the water after t seconds?
Since the vessel is hemispherical, its volume can be given by:V=((4/3)(pi)r3)/2V=(2/3)(pi)r3where r is the radius of the vessel.Since water is flowing into the vessel at a constant rate of 1 L/s, the volume of water in the vessel is thereby increasing at a constant rate of 1 L/s.By deriving the volume equation for the vessel with respect to time, we can equate the rate of change of the volume of water to the rate of change of the radius of the surface of water:dV/dt = (2/3)(pi)(3r2)(dr/dt)You must derive implicitly, so r3 derives down to 3r2(dr/dt) since the radius is also in itself a function of time. This equation can be cleaned up:dV/dt = 2(pi)r2(dr/dt)By solving for dr/dt, we get an expression for rate of change of the radius of the surface of water.dr/dt = (dV/dt)/(2(pi)r2)From the problem, we know that dV/dt is 1 L/s, and the radius of the hemisphere is a constant 0.5 m. We can substitute these known values into the equation:dr/dt = 1/(2(pi)(0.5)2)dr/dt = 2/piThis is the rate of change of the radius of the surface of water. The rate of change is a constant, which is important. Since it is constant, you can simply multiply this rate of change by a quantity of time to find the radius of the water level at any specific time. This is analogous to multiplying a constant velocity times a quantity of time to know an object's position at that time (a rate of change times an amount of change). We know that the vessel has an overall radius of 0.5 m, so the radius of the surface of water cannot exceed 0.5m.(dr/dt)= 2/pitherefore, depth at time t, D= 2t/piThis model gives the depth of the water (D) at any given time (t). As t increases, D(t) will return larger and larger values, which is expected since the water depth will increase as more water flows in.
