Find dy/dx and d2y/dx2. x = et, y = te−t

1 answer:

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Below is the solution:

x = e^t
y = te^-t 

dy/dx = y'/x' 
= (1-t)e^-t/e^t 
= (1-t)e^-2t 

d^2/dx^2 = (x'y"-x"y')/x'^3 
= ((e^t)(t-2)e^-t - (e^t)(1-t)e^-t)/e^3t 
= (2t-3)e^-3t 

or, do it directly 

x = e^t, so 
y = lnx/x 

y' = (1-lnx)/x^2 = (1-t)e^-2t 
y" = (2lnx-3)/x^3 = (2t-3)e^-3t

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Triangles Unit Test part 1 A length of rope is stretched between the top edge of a building and a stake in the ground. The head

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Answer:

The building is 32 feet tall.

Step-by-step explanation:

Consider the figure drawn below representing the given scenario.

AB represents the height of building, BC is the distance between the stake and building, C represents the stake at ground level, and DE represents the height of the tree growing halfway between building and stake.

Since the tree is growing halfway between B and C, therefore, the point E divides the line segment BC into 2 equal parts.

Therefore, $BE = EC$ or E is the midpoint of BC.

Also, from the figure, it is clear that both tree and building are vertical to the ground. So, DE || AB.

Now, from converse of mid-segment theorem, if a line passes through the midpoint of one side of a triangle and also parallel to the third side, then the line also passes through the midpoint of the second side and half the length of the third side (parallel side)

So,

$DE=\frac{1}{2}\times AB\\\\16=\frac{AB}{2}\\\\AB=16\times 2=32\ ft$