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Zach V.

Duke Graduate, Current Medical Student

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SAT II Mathematics Level 2

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Question:

How many years will it take a savings account earning 7.3% interest compounded continuously to triple in value?

Zach V.

Answer:

The formula for continuously compounded interest is:$(A = P\mathrm{e}^{rt}$)The tripled value, $$3P$$, can be substituted into the equation for $$A$$.$(3P=P\mathrm{e}^{rt}$).Then, the value for $$t$$ can be solved for.$(t=\frac{\ln(3)}{r}$).This yields an answer:$(r=15.05 \; years$)

MATLAB

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Question:

Write a function that takes two inputs, $$x$$ and $$y$$, plots the relationship between the two data sets, and outputs the $$R^2$$ value of the linear fit.

Zach V.

Answer:

function R2 = LinReg(x, y) n = length(y); VarX = (sum((x-mean(x)).^2)./n)^.5; VarY = (sum((y-mean(y)).^2)./n)^.5; Sum = sum((x-mean(x)).*(y-mean(y))); Ans = Sum./(VarX*VarY); R2 = (Ans/n)^2; figure(1);clf; hold on plot(x, y,'k-'); grid on end

Calculus

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Question:

Consider an inverted conical container of water. The radius, $$r$$, of the water is related to the height, $$h$$, of the water such that: $(r = 80\mathrm{e}^{h}$)If the cone is being filled with water, determine the rate at which the diagonal length of the cone is changing when the rate at which the surface area of the water in the cone is changing at a rate of 100 $$in^2/s$$ and the radius is changing at a rate of 0.03 $$in/s$$.

Zach V.

Answer:

The surface area of the water can be modeled as a circle, since the water is in a cone: $(A = \pi r^2$)The rate at which the area is changing is:$(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$)This can be arranged to solve for the radius, using the rates of change listed in the problem statement:$(r=530.52 \;in$)The diagonal portion of the cone can be modeled as the hypotenuse of a right triangle with side lengths $$r$$ and $$h$$:$(s^2 = r^2 + h^2$)This can be written in terms of just $$s$$ and $$r$$ by rearranging the equation for the radius and substituting:$(r = 80\mathrm{e}^h$)$(h =\ln\left(\frac{r}{80}\right)$)$(s^2 = r^2 + \ln\left(\frac{r}{80}\right)^2$)Now, the derivative of this expression can be taken with respect to time.$(2s\frac{ds}{dt}=2r\frac{dr}{dt}+2\ln\left(\frac{r}{80}\right)\frac{80}{r}\frac{1}{80}\frac{dr}{dt}$).The values of $$r$$ and $$\frac{dr}{dt}$$ can be substituted into the equation. The value for $$s$$ can be calculated with the equation $$s^2 = r^2 + \ln\left(\frac{r}{80}\right)^2$$.This yields a rate of change for the diagonal portion of the cone of:$(\frac{ds}{dt}=0.0000565\;in/s$)

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