Find the equation of the tangent line and normal line to the curve at a given point y=x^4+2e^x

1 answer:

4 0

The slope of the line is given by the derivative

m = dy/dx = 4x^3 + 2e^x

using the point slope form of a straight line
y - y1 = m(x - x1) where m is the above and (x1,y1) is the point of contact of the tangent we have

y - y1 = (4x^3 + 2e^x)(x - x1) which is equation of the tangent

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Aiko stands 5 m from a tree. At that distance, the angle of elevation from the ground to the top of the tree is 80°.

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Answer:

$\boxed {\boxed {\sf About \ 28.36 \ meters }}$

Step-by-step explanation:

Assuming the tree is perpendicular to the ground, we can use the right triangle trigonometric ratios to find the tree's height.

sin(θ)= opposite/hypotenuse
cos(θ)= adjacent/hypotenuse
tan(θ)= opposite/adjacent

Now, let's draw a diagram. We know Aiko is 5 meters from the base of the tree. From there, the angle to the top of the tree is 80 degrees. We are looking for x, the tree's height. The diagram attached is not to scale.

We base the sides off of the angle. x is opposite of 80 degrees and 5 is adjacent. Therefore we must use tangent.

$tan(\Theta)=\frac{opposite}{adjacent}$