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Calculus 2 : Parametric Calculations

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Example Questions

Example Question #1 : Parametric Calculations

Calculate the length of the curve drawn out by the vector function ${\mathbf{v}(t)} = \left \langle t,\sqrt{2}}e^t,\frac{1}{2}e^{2t} \right \rangle,$ from 0<t<1 .

Possible Answers:

$\frac{1}{2}(e^2+1)$

ln(2) -1

e+1

None of the other answers.

e^2 +e^4

Correct answer:

$\frac{1}{2}(e^2+1)$

Explanation:

The formula for arc length of a parametric curve in space is $\int_{a}^{b} \sqrt{(\frac{d\mathbf{v}_1}{dt})^2 +(\frac{d\mathbf{v}_2}{dt})^2 +(\frac{d\mathbf{v}_3}{dt})^2} dt$ for a <t<b .

Taking derivatives of each of the vector function components and substituting the values into this formula gives

$\int_{0}^{1} \sqrt{1 + 2e^{2t} + e^{4t}} dt$

We need to recognize that underneath the square root we have a perfect square, and we can write it as

$\int_{0}^{1} \sqrt{(e^{2t} + 1)^2} dt$

Solving this we get

$\int_{0}^{1} e^{2t}+1 dt = [\frac{1}{2}e^{2t} +t]_0^1$ $= \bigg(\frac{1}{2}e^2+1\bigg)-\bigg(\frac{1}{2}\bigg) = \frac{1}{2}(e^2+1)$

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Example Question #2 : Parametric Calculations

Calculate $\frac{dy}{dx}$ at the point (1,2) on the curve defined by the parametric equations

x(t) = e^t , y(t)=ln(t+1)+t^2 +2

Possible Answers:

ln(2)

None of the other answers

ln(3)

Correct answer:

None of the other answers

Explanation:

The correct answer is .

We use the equation

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{1}{t+1}+2t+0}{e^t}$

But we need a value for to substitute into our derivative. We can obtain such a by setting x(t) =1, y(t) =2 as our given point suggests.

1 = e^t $\Rightarrow t =0$
$2 = ln(t+1)+t^2+2 \Rightarrow t=0$
Since our values of match, t=0 is our correcct value. Substituting this into the derivative and simplifying gives us our answer of

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Example Question #2 : Parametric Calculations

Which of the answers below is the equation obtained by eliminating the parametric from the following set of parametric equations?

x=t^2+3t+5

$y=\frac{3}{2}t-6$

Possible Answers:

$y=\frac{4}{9}x^2+\frac{22}{3}x+33$

$x=\frac{3}{2}y^2+16y+\frac{4}{3}$

$y=-\frac{14}{3}x^2+\frac{4}{7}x-4$

$y=\frac{3}{2}x^2+16x+\frac{4}{3}$

$x=\frac{4}{9}y^2+\frac{22}{3}y+33$

Correct answer:

$x=\frac{4}{9}y^2+\frac{22}{3}y+33$

Explanation:

When the problem asks us to eliminate the parametric, that means we want to somehow get rid of our variable t and be left with an equation that is only in terms of x and y. While the equation for x is a polynomial, making it more difficult to solve for t, we can see that the equation for y can easily be solved for t:

$y=\frac{3}{2}t-6$

$t=\frac{2}{3}y+4$

Now that we have an expression for t that is only in terms of y, we can plug this into our equation for x and simplify, and we will be left with an equation that is only in terms of x and y:

$x=t^2+3t+5=(\frac{2}{3}y+4)^2+3(\frac{2}{3}y+4)+5$

$x=\frac{4}{9}y^2+\frac{16}{3}y+16+2y+12+5$

$x=\frac{4}{9}y^2+\frac{22}{3}y+33$

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Example Question #3 : Parametric Calculations

Suppose x=cos(t) and y=sin(t) . Find the arc length from $0\leq t \leq \pi$ .

Possible Answers:

$\pi$

$\frac{\pi}{2}$

$\sqrt{\pi}$

$\sqrt{\frac{\pi}{2}}$

$2\pi$

Correct answer:

$\pi$

Explanation:

Write the arc length formula for parametric curves.

$L=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\:dt$

Find the derivatives. The bounds are given in the problem statement.

$\frac{dx}{dt}=-sin(t)$

$\frac{dy}{dt}=cos(t)$

$L=\int_{0}^{\pi}\sqrt{(-sin(t))^2+(cos(t))^2}\:dt$

$L=\int_{0}^{\pi}\sqrt{sin(t)^2+cos(t)^2}\:dt$

$L=\int_{0}^{\pi}1\:dt =[t]_{0}^{\pi}= \pi$

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Example Question #5 : Parametric Calculations

Solve for $\frac{dy}{dx}$ if $y=7t^{2}-12$ and $x=5t^{2}+2$ .

Possible Answers:

$-\frac{5}{7}$

$-\frac{7}{5}$

None of the above

$\frac{5}{7}$

$\frac{7}{5}$

Correct answer:

$\frac{7}{5}$

Explanation:

Given equations for and in terms of , we can find the derivative of parametric equations as follows:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ , as the terms will cancel out.

Using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ and given $y=7t^{2}-12$ and $x=5t^{2}+2$ :

$\frac{dy}{dt}=(2)7t^{2-1}-(0)12=14t$

$\frac{dx}{dt}=(2)5t^{2-1}+(0)2=10t$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{14t}{10t}=\frac{7}{5}$ .

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Example Question #3 : Parametric Calculations

Solve for $\frac{dy}{dx}$ if $y=9t^{2}-13$ and $x=t^{2}+15$ .

Possible Answers:

$-\frac{1}{9}$

$\frac{1}{9}$

None of the above

Correct answer:

Explanation:

Given equations for and in terms of , we can find the derivative of parametric equations as follows:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ , as the terms will cancel out.

Using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ and given $y=9t^{2}-13$ and $x=t^{2}+15$ :

$\frac{dy}{dt}=(2)9t^{2-1}-(0)13=18t$

$\frac{dx}{dt}=(2)t^{2-1}+15=2t$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{18t}{2t}=9$

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Example Question #4 : Parametric Calculations

Solve for $\frac{dy}{dx}$ if y=2t-9 and x=5t+10 .

Possible Answers:

None of the above

$\frac{5}{2}$

$\frac{2}{5}$

$-\frac{2}{5}$

$-\frac{5}{2}$

Correct answer:

$\frac{2}{5}$

Explanation:

Since we have two equations y=2t-9 and x=5t+10 , we can find $\frac{dy}{dx}$ by dividing the derivatives of the two equations - thus:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ since the terms cancel out by standard rules of division of fractions.

In order to find the derivatives of and , let's use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ :

$y=2t-9\rightarrow \frac{dy}{dt}=(1)2t^{1-1}-0(9)=2$

$x=5t+10\rightarrow\frac{dx}{dt}=(1)5t^{1-1}+(0)10=5$

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2}{5}$ .

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Example Question #4 : Parametric Calculations

Solve for $\frac{dy}{dx}$ if y=7t+4 and x=7t-8 .

Possible Answers:

$\frac{1}{7}$

Correct answer:

Explanation:

Since we have two equations y=7t+4 and x=7t-8 , we can find $\frac{dy}{dx}$ by dividing the derivatives of the two equations - thus:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ since the terms cancel out by standard rules of division of fractions.

In order to find the derivatives of and , let's use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ :

$y=7t+4\rightarrow \frac{dy}{dt}=(1)7t^{1-1}+(0)4=7$

$x=7t-8\rightarrow\frac{dx}{dt}=(1)7t^{1-1}-(0)8=7$

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{7}{7}=1$ .

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Example Question #1 : Parametric Calculations

Solve for $\frac{dy}{dx}$ if $y=5t^{2}+6t-9$ and $x=2t^{2}-6t+12$ .

Possible Answers:

$-\frac{4}{10}$

$\frac{4}{10}$

None of the above

$\frac{5t+3}{2t-3}$

$\frac{5t-3}{2t-3}$

Correct answer:

$\frac{5t+3}{2t-3}$

Explanation:

Since we have two equations $y=5t^{2}+6t-9$ and $x=2t^{2}-6t+12$ , we can find $\frac{dy}{dx}$ by dividing the derivatives of the two equations - thus:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ (since the terms cancel out by standard rules of division of fractions).

In order to find the derivatives of and , let's use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ :

$y=5t^{2}+6t-9\rightarrow \frac{dy}{dt}=(2)5t^{2-1}+(1)6t^{1-1}-(0)9=10t+6$

$x=2t^{2}-6t-12\rightarrow \frac{dx}{dt}=(2)2t^{2-1}-(1)6t^{1-1}-(0)12=4t-6$

Therefore, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{10t+6}{4t-6}=\frac{2(5t+3)}{2(2t-3)}=\frac{5t+3}{2t-3}$ .

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Example Question #4 : Parametric Calculations

Given x=5t-1 and y=2t+9 , what is the length of the arc from $0\leq t\leq3$ ?

Possible Answers:

$3\sqrt{29}$

$\sqrt{29}$

$2\sqrt{29}$

$4\sqrt{29}$

Correct answer:

$3\sqrt{29}$

Explanation:

In order to find the arc length, we must use the arc length formula for parametric curves:

$L=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt$ .

Given x=5t-1 and y=2t+9 , we can use using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ , to derive

$\frac{dx}{dt}=(1)5t^{1-1}-(0)1=5$ and $\frac{dy}{dt}=(1)2t^{1-1}+(0)9=2$ .

Plugging these values and our boundary values for into the arc length equation, we get:

$L=\int_{0}^{3}\sqrt{(5)^{2}+(2)^{2}}dt$

$L=\int_{0}^{3}(\sqrt{25+4})dt$

$L=\int_{0}^{3}(\sqrt{29})dt$

Now, using the Power Rule for Integrals $\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ , we can determine that:

$L=(t\sqrt{29})_{0}^{3}\textrm{}$

$L=3\sqrt{29}-0\sqrt{29}$

$L=3\sqrt{29}$

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Calculus 2 : Parametric Calculations

Study concepts, example questions & explanations for Calculus 2

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