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For those of you who are struggling to understand vector calculus, this is the blog post for you! I'll walk you through the process, and guarantee that by the end of this essay we will both be experts in vector calculus. Beginning at point (x0, y0), and given a velocity v: x' = x0 + v*t; y' = y0 + v*t; z' = 0; Where t is time. It can also be said as x(t) = x(xo, yo) + vxt and y(t) = y(yo, yo). The length of the vector is given by the magnitude of the vector: |v| = "sqrt(vx^2 + vy^2)", and will be used to solve for velocity. Vectors can be added to one another, and substracted from one another, so long as they have the same length. This is why, when solving for velocity, we need to find the length of the vector. We can find this by simply drawing out the graph of both vectors. This is what we get when we add vectors: The length of this vector is given in the equation , where b = y(yo, yo) - x(xo, yo). And in simpler terms it's (x' - xo)- (y' - yo). The dot product of two vectors gives you their magnitude times their direction, and can be defined as: where a and b are two different vectors. This means that v + v' = v|v| + v|v|2 = ab. Since we already know the magnitude of each vector, we can easily find v, or v' for that matter. Looking at the diagram above, we see that: x(xo - 3*t)/(t + 2) = x0 - 3/5 * t; y(yo + 1.5 * t- 3.75)/(t + 2) = yo + 1/6 * t; And substituting these into the equation results in: 0+ 3*1/5 - 6/6 + 4/(t+2) - 5/3 = 0. Converting this into an equation such as: 4 = 0 + 3*t - 6/6 + 4/(t+2) - 5/3 We find that t = 1.41 time units. We then solve for the velocity: v = (1.41)/(5*6/52) = 0.019. Now that you know the steps, here's an example of how to use vector calculus to solve a problem. cfa1e77820
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