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**Linear Approximation and Differentials**

Lesson 3.9

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**Tangent Line Approximation**

Consider a tangent to a function at a point where x = a Close to the point, the tangent line is an approximation for f(x) y=f(x) The equation of the tangent line: y = f(a) + f ‘(a)(x – a) f(a) • a

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**Tangent Line Approximation**

We claim that This is called linearization of the function at x=a. Recall that when we zoom in on an interval of a differentiable function far enough, it looks like a line.

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**New Look at dy = rise of tangent relative to x = dx**

• • • x x + x x = dx dy = rise of tangent relative to x = dx y = change in y that occurs relative to x = dx

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**New Look at We know that Recall that dy/dx is NOT a quotient**

then Recall that dy/dx is NOT a quotient it is the notation for the derivative However … sometimes it is useful to use dy and dx as actual quantities

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**The Differential of y Consider Then we can say**

this is called the differential of y the notation is d(f(x)) = f ’(x) * dx it is an approximation of the actual change of y for a small change of x

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**Animated Graphical View**

Note how the "del y" and the dy in the figure get closer and closer

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**Differentials Example: Find the differential of the function:**

Remember a differential is dy = f ‘(x)· dx (assuming y=f(x))

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**Try It Out Note the examples and rules for differentials on page 238.**

Find the differential of: 1) y = 3 – 5x2 2) f(x) = xe-2x

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**Differentials Example Finding the Differential dy**

Find the differential dy and evaluate dy for the given value of x and dx.

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**Differential Estimate of Change (needed for #34 and #37 on the homework)**

Three types of changes that can be found Absolute, Relative and Percent Change Absolute True Estimated Relative and Percent Change True Estimate Relative Percent % %

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**<- Approximate error in volume**

Differentials Example: A company makes ball-bearings with a radius of 2 inches. The measurements are considered to be correct to within 0.01 in. Use differentials to determine the approximate error in volume, approximate relative error in volume, AND approximate relative error percentage in volume. To solve this problem, we will use the equation: ∆V ≈ dV = V ‘(r)· dr <- Approximate error in volume

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Differential Example: A company makes ball-bearings with a radius of 2 inches. The measurements are considered to be correct to within 0.01 in. Use differentials to determine the approximate error in volume, approximate relative error in volume, AND approximate relative error percentage in volume. The estimate of relative error is given by: <- Approximate relative error in volume The estimate percentage of relative error is given by: <- Approximate relative error percentage in volume

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**Differentials for Approximations**

Consider Use Then with x = 25, dx = .3 obtain approximation

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**Propagated Error Consider a rectangular box with a square base**

Height is 2 times length of sides of base Given that x = 3.5 You are able to measure with 3% accuracy What is the error propagated for the volume? x 2x x

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**Propagated Error We know that**

Then dy = 6x2 dx = 6 * 3.52 * = This is the approximate propagated error for the volume

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**Propagated Error The propagated error is the dy The relative error is**

sometimes called the df The relative error is The percentage of error relative error * 100% (in this case, it would be 9%)

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DERIVATIVES 3. Summary f(x) ≈ f(a) + f’(a)(x – a) L(x) = f(a) + f’(a)(x – a) ∆y = f(x + ∆x) – f(x) dx = ∆x dy = f’(x)dx ∆ y≈ dy.

DERIVATIVES 3. Summary f(x) ≈ f(a) + f’(a)(x – a) L(x) = f(a) + f’(a)(x – a) ∆y = f(x + ∆x) – f(x) dx = ∆x dy = f’(x)dx ∆ y≈ dy.

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