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Calculus 2 : Derivatives of Parametrics

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Example Questions

Example Question #1 : Derivatives Of Parametrics

Find the derivative of the following set of parametric equations:

x=2t^3+5t^2-11t+3

y=7t^2-4t-18

Possible Answers:

$\frac{t^2-7t+8}{6t-2}$

$\frac{2(7t-2)}{6t^2+10t-11}$

$\frac{5t^2+6t+3}{2t-11}$

$\frac{-14t+1}{8t^2+5t-2}$

$\frac{4(3t+4)}{2t^2+4t-3}$

Correct answer:

$\frac{2(7t-2)}{6t^2+10t-11}$

Explanation:

We start by taking the derivative of x and y with respect to t, as both of the equations are only in terms of this variable:

$\frac{dx}{dt}=6t^2+10t-11$

$\frac{dy}{dt}=14t-4$

The problem asks us to find the derivative of the parametric equations, dy/dx, and we can see from the work below that the dt term is cancelled when we divide dy/dt by dx/dt, leaving us with dy/dx:

$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dt}\frac{dt}{dx}=\frac{dy}{dx}$

So now that we know dx/dt and dy/dt, all we must do to find the derivative of our parametric equations is divide dy/dt by dx/dt:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{14t-4}{6t^2+10t-11}=\frac{2(7t-2)}{6t^2+10t-11}$

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Example Question #2 : Derivatives Of Parametrics

Solve:

$\int sin^3xcos^4x dx$

Possible Answers:

$\frac{1}{4} cos^4x - \frac{1}{6} cos^6x + C$

$\frac{1}{5} sin^5x - \frac{1}{6} sin^6x + C$

$-\frac{1}{5} cos^5x + \frac{1}{7} cos^7x + C$

$\frac{1}{4} sin^4x + \frac{1}{7} sin^8x + C$

$\frac{1}{5} sin^5x - \frac{1}{7} cos^5x + C$

Correct answer:

$-\frac{1}{5} cos^5x + \frac{1}{7} cos^7x + C$

Explanation:

The integration involves breaking up a power of a trigonometric ratio, and then using known trigonometric identities.

$\int sin^3x cos^4x dx = \int [sin x \cdot sin^2x \cdot cos^4x]dx$ $= \int \left[sin x \cdot \left(\frac{1}{2} - \frac{1}{2}cos 2x\right)\cdot cos^4x\right] dx$

$= \frac{1}{2}\int [sin x cos^4x - sin x cos 2x cos^4x]dx$

$= \frac{1}{2} \int [sin x cos^4x - sin x (2cos^2x - 1)cos^4x ]dx$

$= \frac{1}{2} \int [2sin x cos^4x - 2sin x cos^6x]dx$

$= \frac{1}{2} \cdot -\frac{1}{5} \cdot 2cos^5x - \frac{1}{2} \cdot 2 \cdot- \frac{1}{7}cos^7x + C$

$= -\frac{1}{5} cos^5x + \frac{1}{7} cos^7x + C$

The alternative is to find which answer choice has a derivative equal to the answer choice, and for this we get:

${}\frac{\mathrm{d} }{\mathrm{d} x}\left(-\frac{1}{5} cos^5x + \frac{1}{7} cos^7x + C\right) = cos^4x\cdot sin(x) - cos^6x\cdot sin(x)$

${}= cos^4xsin(x)[1 - cos^2x] = cos^4x\cdot sin(x)\cdot sin^2x = sin^3xcos^4x$

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Example Question #1 : Derivatives Of Parametrics

Solve for $\frac{dy}{dx}$ if x=2t^3 and y=3t^4 .

Possible Answers:

$\frac{3}{2}$

$\frac{3}{2}t$

Correct answer:

Explanation:

Write the the formula to solve for the derivative of parametric functions.

$\large \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Find $\frac{dy}{dt}$ and $\frac{dx}{dt}$ using the power rule $f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

$\frac{dy}{dt}=12t^3$

$\frac{dx}{dt}=6t^2$

Substitute back to the formula to obtain the derivative.

${} \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{12t^3}{6t^2}=2t$

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Example Question #2 : Derivatives Of Parametrics

Find the derivative of the following parametric function:

$x=t^2-1, y=\cos(t)+\sin(t)$

Possible Answers:

$\frac{2t}{\cos(t)-\sin(t)}$

$\frac{\cos(t)-\sin(t)}{2t}$

$\frac{\sin(t)-\cos(t)}{2t}$

$\cos(t)-\sin(t)$

Correct answer:

$\frac{\cos(t)-\sin(t)}{2t}$

Explanation:

The derivative of a parametric function is given by:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

where

$\frac{dy}{dt}=\cos(t)-\sin(t)$ , $\frac{dx}{dt}=2t$

The derivatives were found using the following rules:

$\frac{d}{dx}\cos(x)=-\sin(x)$

$\frac{d}{dx}\sin(x)=\cos(x)$

$\frac{d}{dx}(x^n)nx^{n-1}$

Simply divide the derivatives as shown above.

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Example Question #1 : Derivatives Of Parametrics

Solve for $\frac{dy}{dx}$ if $y=3t^{2}-2$ and $x=4t^{2}+6$ .

Possible Answers:

$\frac{4}{3}$

$\frac{4}{3}t$

$\frac{3}{4}t$

$\frac{3}{4}$

None of the above

Correct answer:

$\frac{3}{4}$

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 $y=3t^{2}-2$ and $x=4t^{2}+6$ ,

$\frac{dy}{dt}=(2)3t^{2-1}-(0)2=6t$ and $\frac{dx}{dt}=(2)4t^{2-1}+(0)6=8t$ .

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{6t}{8t}=\frac{3}{4}$ .

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Example Question #1 : Derivatives Of Parametrics

Find the derivative of the following parametric equation:

$x=2t+\cos(t), y=t^2$

Possible Answers:

$\frac{2t}{2-\sin(t)}$

$\frac{2t}{2+\sin(t)}$

$\frac{2+\sin(t)}{2t}$

$\frac{2-\sin(t)}{2t}$

Correct answer:

$\frac{2t}{2-\sin(t)}$

Explanation:

The derivative of a parametric equation is given by the following equation:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Solving for the derivative of the equation for y, you get

$\frac{dy}{dt}=2t$

and for the equation for x, you get

$\frac{dx}{dt}=2-\sin(t)$

The following rules were used for the derivatives:

$\frac{d}{dx}(x^n)=nx^{n-1}$ , $\frac{d}{dx}\cos(x)=-\sin(x)$

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Example Question #6 : Derivatives Of Parametrics

Find $\frac{dx}{dy}$ if x=cos(t) and y=sin(t) .

Possible Answers:

cot(t)

-cot(t)

-tan(t)

sec(t)

tan(x)

Correct answer:

-tan(t)

Explanation:

Write the formula to find the derivative for parametric equations.

$\frac{dx}{dy}=\frac{\frac{dx}{dt}}{\frac{dy}{dt}}$

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

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

Substitute the knowns into the formula.

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

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Example Question #7 : Derivatives Of Parametrics

Solve for $\frac{dy}{dx}$ if x=5t+9 and y=2t+7 .

Possible Answers:

$-\frac{5}{2}$

$\frac{2}{5}$

$\frac{5}{2}$

$-\frac{2}{5}$

Correct answer:

$\frac{2}{5}$

Explanation:

We can determine that $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ since the terms will cancel out in the division process.

Since x=5t+9 and y=2t+7 , we can use the Power Rule

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

$\frac{dx}{dt}=(1)5x^{1-1}=5$ and $\frac{dy}{dt}=(1)2t^{1-1}+(0)7=2$ .

Thus:

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

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Example Question #2 : Derivatives Of Parametric, Polar, And Vector Functions

Solve for $\frac{dy}{dx}$ if x=2t-5 and y=3t+8 .

Possible Answers:

$\frac{2}{3}$

$-\frac{2}{3}$

$-\frac{3}{2}$

None of the above

$\frac{3}{2}$

Correct answer:

$\frac{3}{2}$

Explanation:

We can determine that $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ since the terms will cancel out in the division process.

Since x=2t-5 and y=3t+8 , we can use the Power Rule

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

$\frac{dx}{dt}=(1)2t^{1-1}-(0)5=2$ and $\frac{dy}{dt}=(1)3t^{1-1}+(0)8=3$ .

Thus:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3}{2}}$ .

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Example Question #7 : Derivatives Of Parametrics

Solve for $\frac{dy}{dx}$ if x=9t+1 and y=4t+10 .

Possible Answers:

$-\frac{9}{4}$

$\frac{4}{9}$

$\frac{9}{4}$

$-\frac{4}{9}$

None of the above

Correct answer:

$\frac{4}{9}$

Explanation:

We can determine that $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ since the terms will cancel out in the division process.

Since x=9t+1 and y=4t+10 , we can use the Power Rule

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

$\frac{dx}{dt}=(1)9t^{1-1}+(0)1=9$ and $\frac{dy}{dt}=(1)4t^{1-1}+(0)10=4$ .

Thus:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{4}{9}}$ .

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Calculus 2 : Derivatives of Parametrics

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Derivatives Of Parametrics

Example Question #2 : Derivatives Of Parametrics

Example Question #1 : Derivatives Of Parametrics

Example Question #2 : Derivatives Of Parametrics

Example Question #1 : Derivatives Of Parametrics

Example Question #1 : Derivatives Of Parametrics

Example Question #6 : Derivatives Of Parametrics

Example Question #7 : Derivatives Of Parametrics

Example Question #2 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #7 : Derivatives Of Parametrics

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