Differential Equations Exercises

Answer

(a) Let's find the tangent line. First we find the value of f when x = 1:

f (1) = 1³ + 2(1) + 1 = 4

so our point is (1, 4). Now we find the slope. Since f '(x) = 3x² + 2, the slope at the point (1, 4) is

f '(1) = 3(1)² + 2 = 5.

Putting this together, the tangent line is

y = 5x – 1.

For a sanity check, we can put this into our calculator and make sure it looks like a tangent line, which it does.

(b) We want to estimate f (1 + .2), so we put x = 1.2 into the equation for the tangent line:

5(1.2) – 1 = 6 – 1 = 5

Answer

(a) The function value at is

The derivative function is

f '(x) = 2(sin x)(cos x)

so the slope of our tangent line is

Putting this together, the tangent line is

(b) To estimate , we plug this value into the tangent line:

Answer

(a) When x = 8 we have

f (8) = log₂ 8 = 3.

The derivative of f is

so the slope of the tangent line is

This means the equation for the tangent line is

(b) We plug 8.01 into the tangent line equation:

Answer

(a) We know f (0) = 4(0) + e⁰ = 1, so the point we need for the tangent line is (0, 1). The derivative of f is 4 + e^x, so the slope of the tangent line is

f '(0) = 4 + e⁰ = 5.

Putting this together, the equation for the tangent line is

y = 5x + 1

(b) Plug 0 + 0.05 = 0.05 into the tangent line equation:

5(0.05) + 1 = 1.25

Answer

(a) The slope of the tangent line at is

The value of the function at is 1, so we get the following tangent line:

(b) We plug

into the tangent line and get

It turns out that when you know a value of f at a particular x and you're asked to approximate the value of f somewhere nearby, you don't need to go to the trouble of finding the tangent line equation. As we saw earlier, it's enough to know the slope of the tangent line, y_old, and Δ x. That's enough information to find y_new.

Answer

We have y_old = 4.2, Δ x = 0.1, and slope = 0.15. So

Answer

We have , slope  , and . So

This is believable. We expect y_old and y_new to be close together, and they are:

Answer

We have

We need to know f '(-1), also known as the slope of the tangent line. Since we have two points on the line we can find this.

The slope of the line is

Now we can find the the tangent line:

We can find b by plugging in either point given into the line.

We'll use (-4, 5.5):

so

Now that we have the tangent line, we can estimate f(-0.9):

which is about 3.43.

Answer

We know that f '(3) is the same as the slope of the tangent line to f at x = 3, so all we need to do is find the slope of this line.

We get

Answer

We have f (2) = 7. A tangent line approximation at x = 2 estimates f (2.1) ≈ 7.2. This means if we draw the tangent line to f at 2, the point (2.1,7.2) is on that line.

We know that f '(2) is the same as the slope of the tangent line to f at x = 2, so we need to find the slope of this tangent line. The slope is

so we conclude

f '(2) = 2.

Answer

Let's take the tangent line approximation at x = 2, since that's a nice round number that's close to 2.01. Then

Δ x = 0.01

y_old = 3(2)² – 4(2) + 1 = 5.

Since f '(x) = 6x – 4, the slope of the tangent line at x = 2 is

f '(2) = 6(2) – 4 = 8.

therefore

Answer

Since 2.99 is close to 3, we'll take the tangent line to f at 3. Then Δ x = -0.01 and

y_old = f (3) = 7(9) – 3(3) + 4 = 58.

The derivative of f is

f '(x) = 14x – 3

so the slope of the tangent line at 3 is

f '(3) = 14(3) – 3 = 39.

We can now estimate f (2.99) by finding y_new on the tangent line:

Answer

Let's take the tangent line at x = 0, since that's close to 0.05. Then Δ x = 0.05.

Since f '(x) = f (x) = e^x, both y_old and the slope of the tangent line equal e⁰ = 1.

Then

Answer

Let's take the tangent line to f at x = 3, which is close to 3.02. Then Δ x = .02. We have

y_old = f (3) = 2³ = 8.

The derivative of f is

f '(x) = 2^x ln 2

so the slope of the tangent line at 3 is

f '(3) = 8 ln 2.

Now we can do the estimation:

We didn't round until the very last step.

Answer

Following the hint, let's draw the tangent line at x = π, which is close to 3. Then

Δ x = 3 – π.

If we're drawing a tangent line at π to approximate f (3), then yes, Δ x will be negative.

We'll wait to round until the very end of the problem. Now let's figure out the other values we need.

y_old = f (π) = sin(5π) = 0.

Since

f '(x) = 5cos(5x)

the slope of the tangent line is

f '(π) = 5cos(5π) = -5.

Now we can do the estimation, leaving all rounding until the end:

Answer

We have y_old = 2, Δ x = 0.5, and

So

Answer

We have y_old = 3, Δ x = -.5, and

So

Answer

We have y_old = 5, Δ x = .01, and

So

Answer

We have y_old = 3, Δ x = -.05, and

So

Answer

We have y_old = 5, Δ x = .01, and

So