Hayley G.

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Pre-Calculus

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Question:

Find an equation of the line which passes through the point $(4,-7)$ and has slope 3.

Hayley G.

Answer:

Recall the Point-Slope Form: The line through the point $(x_1,y_1)$ with slope m has the equation $y-y_1=m(x-x_1)$. Using the Point-Slope Form, we obtain the equation $y-(-7)=3 (x-4)$ or, simplifying, $y+7=3 (x-4)$ or $y=3x-19$.

Numerical Analysis

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Question:

Find the local truncation error of forward Euler method

Hayley G.

Answer:

The local truncation error is defined as: LTE=$$\frac{1}{h}(y(t_{n+1})-y_{n+1})$$ where $y(t_{n+1})$ is the exact value and $y_{n+1}$ the approximate value. Recall that Forward Euler method has the form: $$y_{n+1}=y_n+hy_n'$$ Taylor expand the exact value $y(t_{n+1})$. $$y(t_{n+1})=y(t_n)+hy'(t_n) +O(h^2)$$ Replace all approximations (e.g. y_n) on the right-hand side of the method with their exact counterparts (eg. y(t_n)) so we have for Eulers method: $$y_{n+1}=y(t_n)+hy'(t_n)$$ Using the formula for LTE and substituting our values for $y(t_{n+1})$ and $y_{n+1}$ we obtain: $$LTE= \frac{1}{h}(y(t_n)+hy'(t_n)+O(h^2)-y(t_n)-hy'(t_n)) =\frac{1}{h}O(h^2) =O(h)$$ The local truncation error is O(h).

Calculus

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Question:

Use the limit definition to compute the derivative, $f'(x)$, for $f(x)=\frac{1}{2}x-\frac{3}{5}$

Hayley G.

Answer:

The limit definition of the derivative is given by: $$f'(x)=lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ We know $f(x)=\frac{1}{2}x-\frac{3}{5}$. Now to calculate $f(x+h)$ we replace x by x+h in the definition of $f(x)$. So $$f(x+h)=\frac{1}{2}(x+h)-\frac{3}{5}= \frac{1}{2}x+\frac{1}{2}h-\frac{3}{5}$$ Combining these parts into the limit definition of the derivative we obtain: $$f'(x)=lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ $$=lim_{h \rightarrow 0) \frac{\frac{1}{2}x+\frac{1}{2}h-\frac{3}{5}-\frac{1}{2}x+\frac{3}{5}}{h}$$. Simplifying and taking a limit yields: $$f'(x)=\frac{1}{2}$$

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