Let f(x)=\sqrt{100 - x^2}. Use differentials (linear approximation) to estimate f(6.2). Base your approximatio

Question

Give your answer in at least 4 signficant figures or in a fraction.

Answer 1

Let f(x)= y = √{100 - x²}.

dy =½ * -2x/√{100 - x²}. dx
= -x//√{100 - x²}. dx

So δy ≈ -x//√{100 - x²}.δx

f(6) = √{100 - x²}
= √{100 - 6²}
= 8

f(6.2) = f (6 + 0.2)
≈f (6) + -6/8 * 0.2
= 8 - 0.150
= 7.850

Check

f(6.2) = √{100 - (6.2)²}
= √{100 - 38.44}
= √{61.56}
= 7.846 ......

Answer 2

We know that dy=f'(x)dx, and we can calculate f(6)=8.
Differentiating the equation:
f'(x)=(1/2) [(100-x^2)^(-1/2)] (-2x)
Simplifying:
-x [(100-x^2)^(-1/2)]
Plugging in x=6 we get.
-6[100-36]^(-1/2)] = -6(1/8) =-3/4

we know dx=.2, multiplying, we get:
dy=(-0.75)*(.2)=-0.15
Add the value of f(6) to get:
f(6.2)≈7.85

Answer 3

(x+h) ^ 2 = x^2 + 2hx + h^2.
for h small the above is x^2 +2hx

so 100 - x^2 , and x = 6.2
this is 100 -36 +2*0.6*2 = 64 + 2.4 = 66.4

sqrt(x+h) = sqrt(x) + 0.5h ; for h small

thus sqrt(66.4) = 8 + 0.5*2.4 = 9.2 ( which is way too large )sniff.