1, x = t

^{2 }+ 5t +4 , y= 4t , t=o2, x = cosθ, y = 3sinθ , θ=0

1, x = t^{2 }+ 5t +4 , y= 4t , t=o

2, x = cosθ, y = 3sinθ , θ=0

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For 1) I do not think that there is an easy way to solve without a little calculus.

In the calculus approach, we differentiate each equation with respect to t to obtain

dx/dt = 2t +5 and dy/dt = 4. dividing the second of these by the first we get

dy/dx = the slope = 4/(2t +5). for t = 0 this evaluates to 4/5.

To find the concavity, we study the form of the curve obtained by eliminating t between the original two original equations. It is easy to see that the curve is a "sideways" parabola opening to the right. By graphing , we can see that the vertex is in the third quadrant. Thus for y = 0 we are at a point on the upper part (branch) of the curve. For this branch, the curve is concave downwards. That is to say the second derivative of y with respect to x is negative.

For 2) We can use the fact that for small θ sinθ ~ θ, and cosθ ~ 1- (1/2) θ^{2} .

So for θ near zero x = 1 - (1/2) θ^{2} and y = 3 θ . Eliminating θ we see that this again is a "sideways" parabola (for small θ) opening this time to the left. The vertex is at (1,0) . The point corresponding to θ = 0 is the vertex itself. The slope of a "sideways" parabola is undefined (aka infinite) at the vertex point. Thus the slope is undefined. Strictly speaking, the concavity is also undefined, because the second derivative is undefined. However, it might be said the the curve is concave toward the left.

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