The surface area, S, of a sphere of radius r feet is S = S(r) = 4πr^2. Find the instantaneous rate of change of the surface area
1 answer:
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Answer:
a) f(x) = 4 - x²
The linear approximation of the function at a=1 is
L(x) = 5 - 2x at a = 1
b) The graph of the function and the linear approximation at that point is attached to this solution.
The curve represent the real function,
f(x) = 4 - x²
The straight line represents the linear approximation of the function at a=1.
L(x) = 5 - 2x
The curve and the function evidently cross paths at x=1 and understandably so.
c) Using the linear approximation obtained at a = 1.
f(1.1) = 2.8
Using the actual function, the actual value of f(1.1) = 2.79
d) Percent error = 0.358%
Step-by-step explanation:
f(x) = 4 - x²
a) The linear approximation of the function at the given point is given as
L(x) = f(a) + f'(a) [x - a]
f(x) = 4 - x²
a = 1
f(a) = 4 - 1² = 3
f'(x) = -2x
f'(a) = -2(1) = -2
L(x) = f(a) + f'(a) [x - a]
L(x) = 3 + (-2)(x - 1)
L(x) = 3 -2x + 2
L(x) = 5 - 2x
L(x) = -2x + 5
f(x) = 4 - x²
L(x) = 5 - 2x at a = 1
b) The graph of the function and the linear approximation at that point is attached to this solution.
The curve represent the real function,
f(x) = 4 - x²
The straight line represents the linear approximation of the function at a=1.
L(x) = 5 - 2x
The curve and the function evidently cross paths at x=1 and understandably so.
c) Use the linear approx. to estimate the given fxn value.
f(1.1)
L(x) = 5 - 2x
L(1.1) = 5 - 2(1.1) = 2.8
Using the function, the actual value of f(1.1) = 4 - 1.1² = 2.79
d) Compute the percent error in your approximation, 100*Iapprox-exactI/IexactI, where the exact value is given by a calculator
Percent error
= 100% × (|approx - exact|)/exact
Approximated value = 2.8
Exact value = 2.79
Percent error = 100% × (2.8-2.79)/2.79
Percent error = 0.358%
Hope this Helps!!!
Here's our equation.
Let's reorder this so t² comes before t.
We want to find out when the ball hits the ground. Essentially, <em>what is t when h=0?</em>
Let's plug in 0 for h and solve.
From here we can use a number of methods to solve.
Methods such as splitting the middle, quadratic formula, etc. are not necessary for this equation, however. The best method here is just plain factoring.
-16t² and 176t are both divisible by 16t.
Now we want t to equal zero. If either of those factors were zero, then the whole thing would be zero when we multiply by it. The first option is obvious. If t were 0, 16t would be 0. That's when the ball is first kicked off the ground. The other, 11-t, works out to be t = 11.
Let's reorder this so t² comes before t.
We want to find out when the ball hits the ground. Essentially, <em>what is t when h=0?</em>
Let's plug in 0 for h and solve.
From here we can use a number of methods to solve.
Methods such as splitting the middle, quadratic formula, etc. are not necessary for this equation, however. The best method here is just plain factoring.
-16t² and 176t are both divisible by 16t.
Now we want t to equal zero. If either of those factors were zero, then the whole thing would be zero when we multiply by it. The first option is obvious. If t were 0, 16t would be 0. That's when the ball is first kicked off the ground. The other, 11-t, works out to be t = 11.
1. We need to get rid of those pesky parentheses. We do this by distributing the 3.
2. We need to get "n" by itself, so we add 9 to both sides.
3. Add like terms.
4. Now, we need to get rid of the -9, so we divide both sides by -9. Remember that dividing by negative numbers changes the direction of the "greater than or equal to" symbol.
5. We divide, and the negatives cancel.
I wrote the answer both ways so that I could tell what the number line needed to look like.
2. We need to get "n" by itself, so we add 9 to both sides.
3. Add like terms.
4. Now, we need to get rid of the -9, so we divide both sides by -9. Remember that dividing by negative numbers changes the direction of the "greater than or equal to" symbol.
5. We divide, and the negatives cancel.
I wrote the answer both ways so that I could tell what the number line needed to look like.