Defining and Differentiating Parametric Equations - AP Calculus BC

For $x(t)=\sin t+\sin 2t$ and $y(t)=\cos t+\cos 2t$ , find $\frac{dy}{dx}$ in terms of $t$ .
Answer: $\dfrac{-\sin t-2\sin 2t}{\cos t+2\cos 2t}$ This parametric differentiation requires dy/dx = (dy/dt)/(dx/dt). For x(t) = sin t + sin 2t, we find dx/dt = cos t + 2cos 2t. For y(t) = cos t + cos 2t, we get dy/dt = -sin t - 2sin 2t. The parametric derivative formula gives dy/dx = (-sin t - 2sin 2t)/(cos t + 2cos 2t). Choice B incorrectly inverts the fraction. The reliable parametric derivative technique is to differentiate each component with respect to t, then form the ratio dy/dt over dx/dt.
For $x(t)=\sin t+t$ and $y(t)=\cos t-t$ , find $\dfrac{dy}{dx}$ .
Answer: $\dfrac{-\sin t-1}{\cos t+1}$ This problem involves parametric differentiation, where we find the derivative dy/dx for curves defined by parametric equations x(t) and y(t). To compute dy/dx using the chain rule, recognize that dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0. First, differentiate x(t) = sin t + t to get dx/dt = cos t + 1, and y(t) = cos t - t to get dy/dt = -sin t - 1. Then, divide these derivatives to obtain dy/dx = (-sin t - 1) / (cos t + 1). A tempting distractor might be (cos t + 1) / (-sin t - 1), which fails because it incorrectly inverts the ratio of the derivatives. Always remember the general strategy for parametric derivatives: compute dy/dt and dx/dt separately, then form their quotient for dy/dx.
A curve is defined by $x(t)=t+\dfrac{1}{t}$ and $y(t)=t^2$ for $t\ne0$ ; find $\dfrac{dy}{dx}$ .
Answer: $\dfrac{2t}{1-\frac{1}{t^2}}$ This question involves parametric differentiation with a rational expression for x(t). Using dy/dx = (dy/dt)/(dx/dt), we first find the derivatives: dy/dt = 2t and dx/dt = 1 - 1/t². Therefore, dy/dx = 2t/(1 - 1/t²), which is choice A. Choice B shows the reciprocal (1 - 1/t²)/(2t), a common error when students confuse the order of division. The key to parametric differentiation is maintaining the correct quotient structure: always y-derivative over x-derivative.
If $x(t)=1+t^2$ and $y(t)=\dfrac{1}{1+t}$ , what is $\dfrac{dy}{dx}$ at $t=1$ ?
Answer: $\dfrac{-1/(1+t)^2}{2t}\Big|_{t=1}=-\dfrac{1}{8}$ To find dy/dx for parametric equations, we use the chain rule formula dy/dx = (dy/dt)/(dx/dt). Given x(t) = 1 + t² and y(t) = 1/(1 + t), we differentiate: dx/dt = 2t and dy/dt = -1/(1 + t)² using the chain rule. Therefore, dy/dx = [-1/(1 + t)²]/(2t). At t = 1, this becomes [-1/(1 + 1)²]/(2·1) = [-1/4]/2 = -1/8. Choice D incorrectly uses -1/(1 + t) instead of -1/(1 + t)² for the derivative of y. When differentiating compositions like 1/(1 + t), always apply the chain rule to get the correct power in the denominator.
A particle follows $x(t)=\frac{t}{2}$ and $y(t)=\sin t$ ; find $\frac{dy}{dx}$ in terms of $t$ .
Answer: $\dfrac{\cos t}{1/2}$ This parametric curve requires dy/dx = (dy/dt)/(dx/dt). For x(t) = t/2, we find dx/dt = 1/2. For y(t) = sin t, we get dy/dt = cos t. The parametric derivative formula yields dy/dx = (cos t)/(1/2) = 2cos t. Choice A shows the unreduced form. The standard parametric differentiation method involves finding derivatives of both components, then dividing dy/dt by dx/dt.
For $x(t)=\arctan t$ and $y(t)=\ln(1+t^2)$ , compute $\frac{dy}{dx}$ in terms of $t$ .
Answer: $\dfrac{2t/(1+t^2)}{1/(1+t^2)}$ This parametric differentiation requires dy/dx = (dy/dt)/(dx/dt). For x(t) = arctan t, we have dx/dt = 1/(1+t²). For y(t) = ln(1+t²), we get dy/dt = 2t/(1+t²). Applying the chain rule gives dy/dx = [2t/(1+t²)]/[1/(1+t²)] = 2t. Choice A shows the unreduced form which simplifies to 2t. The essential parametric derivative technique is to differentiate each component with respect to the parameter, then form their quotient.
A moving point has $x(t)=\sec t$ and $y(t)=\tan t$ ; find $\frac{dy}{dx}$ in terms of $t$ .
Answer: $\dfrac{\sec^2 t}{\sec t\tan t}$ This parametric curve uses the chain rule dy/dx = (dy/dt)/(dx/dt). For x(t) = sec t, we find dx/dt = sec t tan t. For y(t) = tan t, we get dy/dt = sec² t. Applying the parametric derivative formula yields dy/dx = (sec² t)/(sec t tan t) = sec t/tan t = 1/sin t. Choice B incorrectly inverts the fraction. The standard parametric differentiation technique involves finding both derivatives with respect to t, then forming their proper ratio.
Given $x(t)=\sin(2t)$ and $y(t)=\cos(3t)$ , compute $\frac{dy}{dx}$ in terms of $t$ .
Answer: $\dfrac{-3\sin(3t)}{2\cos(2t)}$ This parametric differentiation uses the chain rule dy/dx = (dy/dt)/(dx/dt). For x(t) = sin(2t), we find dx/dt = 2cos(2t). For y(t) = cos(3t), we get dy/dt = -3sin(3t). The parametric derivative formula gives dy/dx = (-3sin(3t))/(2cos(2t)). Choice B incorrectly inverts the fraction by placing dx/dt in the numerator. The key parametric differentiation strategy is to compute both component derivatives, then form the ratio dy/dt over dx/dt.
A particle moves with $x(t)=t^2-3t$ and $y(t)=4t+\sin t$ ; find $\dfrac{dy}{dx}$ .
Answer: $\dfrac{4+\cos t}{2t-3}$ This problem involves parametric differentiation, where we find the derivative dy/dx for curves defined by parametric equations x(t) and y(t). To compute dy/dx using the chain rule, recognize that dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0. First, differentiate x(t) = t² - 3t to get dx/dt = 2t - 3, and y(t) = 4t + sin t to get dy/dt = 4 + cos t. Then, divide these derivatives to obtain dy/dx = (4 + cos t) / (2t - 3). A tempting distractor might be (2t - 3) / (4 + cos t), which fails because it incorrectly inverts the ratio of the derivatives. Always remember the general strategy for parametric derivatives: compute dy/dt and dx/dt separately, then form their quotient for dy/dx.
A parametric curve has $x(t)=t^2+\tan t$ and $y(t)=t^2+\sec t$ ; find $\frac{dy}{dx}$ .
Answer: $\dfrac{2t+\sec t\tan t}{2t+\sec^2 t}$ This parametric curve uses $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ . For $x(t) = t^2 + \tan t$ , we calculate $\frac{dx}{dt} = 2t + \sec^2 t$ . For $y(t) = t^2 + \sec t$ , we get $\frac{dy}{dt} = 2t + \sec t \tan t$ . Using the parametric derivative gives $\frac{dy}{dx} = \frac{2t + \sec t \tan t}{2t + \sec^2 t}$ . Choice B incorrectly inverts the fraction. The fundamental parametric differentiation technique is to compute derivatives of both components, then divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ .
If $x(t)=\sqrt{t}$ and $y(t)=t^2-1$ for $t>0$ , what is $\dfrac{dy}{dx}$ ?
Answer: $4t\sqrt{t}$ This problem involves parametric differentiation to find dy/dx for the given functions x(t) and y(t). To find dy/dx, we apply the chain rule, which states that dy/dx = (dy/dt) / (dx/dt), assuming dx/dt is not zero. First, compute dx/dt = 1/(2√t) from x(t) = √t, and dy/dt = 2t from y(t) = t² - 1. Therefore, dy/dx = 2t / [1/(2√t)] = 4t√t, which matches choice D. A tempting distractor is choice B, which is the unsimplified form 2t / [1/(2√t)] but is actually equal to D, though the question likely expects the simplified version. In general, for parametric equations, always compute the ratio of the y-derivative to the x-derivative with respect to the parameter to obtain the correct slope.
A parametric curve has $x(t)=2t+\tan t$ and $y(t)=t-\sec t$ . What is $\dfrac{dy}{dx}$ in terms of $t$ ?
Answer: $\dfrac{1-\sec t\tan t}{2+\sec^2 t}$ To find dy/dx for these trigonometric parametric equations, we apply the chain rule dy/dx = (dy/dt)/(dx/dt). With x(t) = 2t + tan t and y(t) = t - sec t, we compute dy/dt = 1 - sec t tan t and dx/dt = 2 + sec² t. Therefore, dy/dx = (1 - sec t tan t)/(2 + sec² t). Choice E incorrectly suggests multiplying these expressions rather than dividing them. For parametric curves, always remember that dy/dx is the quotient of the derivatives, representing the instantaneous rate of change.
A curve is parameterized by $x(t)=2^t$ and $y(t)=t2^t$ ; what is $\frac{dy}{dx}$ in terms of $t$ ?
Answer: $\dfrac{2^t+t2^t\ln 2}{2^t\ln 2}$ This parametric curve uses dy/dx = (dy/dt)/(dx/dt). For x(t) = $2^t$ , we find dx/dt = $2^t$ ln 2. For y(t) = t· $2^t$ , we apply the product rule: dy/dt = $2^t$ + t· $2^t$ ln 2. The parametric derivative formula yields dy/dx = ( $2^t$ + $t·2^t$ ln 2)/( $2^t$ ln 2). Choice B incorrectly inverts the fraction. The fundamental parametric differentiation technique is to compute derivatives of both components, then divide dy/dt by dx/dt.
For the curve $x(t)=\cos t+t$ and $y(t)=\sin t-t^2$ , find $\frac{dy}{dx}$ as a function of $t$ .
Answer: $\dfrac{\cos t-2t}{-\sin t+1}$ This parametric differentiation problem uses the chain rule formula dy/dx = (dy/dt)/(dx/dt). For x(t) = cos t + t, we get dx/dt = -sin t + 1. For y(t) = sin t - t², we get dy/dt = cos t - 2t. Applying the chain rule gives dy/dx = (cos t - 2t)/(-sin t + 1). Choice B incorrectly inverts the fraction by putting dx/dt in the numerator. The fundamental parametric derivative technique is to find both component derivatives, then form their ratio with dy/dt on top.
For $x(t)=\sin t+t$ and $y(t)=\cos t-2t$ , what is $\dfrac{dy}{dx}$ as a function of $t$ ?
Answer: $\dfrac{-\sin t-2}{\cos t+1}$ This problem asks for the derivative dy/dx of parametric equations using the chain rule for parametric differentiation. Given x(t) = sin t + t and y(t) = cos t - 2t, we need dy/dx = (dy/dt)/(dx/dt). Computing the derivatives: dy/dt = -sin t - 2 and dx/dt = cos t + 1. Thus dy/dx = (-sin t - 2)/(cos t + 1). Choice E incorrectly multiplies the derivatives rather than dividing them. Remember that parametric derivatives always involve dividing dy/dt by dx/dt, not multiplying.
Given $x(t)=e^{2t}$ and $y(t)=e^t+1$ , compute $\frac{dy}{dx}$ in terms of $t$ .
Answer: $\dfrac{e^t}{2e^{2t}}$ This parametric differentiation requires $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ . For $x(t) = e^{2t}$ , we find $\frac{dx}{dt} = 2e^{2t}$ . For $y(t) = e^t + 1$ , we get $\frac{dy}{dt} = e^t$ . The parametric derivative formula gives $\frac{dy}{dx} = \frac{e^t}{2e^{2t}} = \frac{1}{2e^t}$ . Choice A shows the unreduced form. The reliable parametric derivative strategy is to differentiate each component separately, then form their proper quotient.
A parametric curve has $x(t)=e^{2t}$ and $y(t)=e^t\cos t$ ; what is $\dfrac{dy}{dx}$ at $t=0$ ?
Answer: $\dfrac{e^t(\cos t-\sin t)}{2e^{2t}}\Big|_{t=0}=\dfrac{1}{2}$ For parametric differentiation, we use dy/dx = (dy/dt)/(dx/dt) where derivatives are taken with respect to the parameter t. Given x(t) = $e^{2t}$ and y(t) = $e^t$ ·cos t, we find dx/dt = $2e^{2t}$ using the chain rule, and dy/dt = $e^t$ ·cos t + $e^t$ ·(-sin t) = $e^t$ (cos t - sin t) using the product rule. Thus dy/dx = $e^t$ (cos t - sin t)/( $2e^{2t}$ ) = (cos t - sin t)/( $2e^t$ ). At t = 0, this becomes (cos 0 - sin 0)/( $2e^0$ ) = (1 - 0)/(2·1) = 1/2. Choice C incorrectly uses cos t + sin t in the numerator instead of cos t - sin t. Always apply the product rule carefully when differentiating products in parametric equations.
Given $x(t)=t^2+1$ and $y(t)=\sqrt{t}$ for $t>0$ , what is $\dfrac{dy}{dx}$ at $t=4$ ?
Answer: $\dfrac{1}{2\sqrt{t}}\Big/2t\Big|_{t=4}=\dfrac{1}{32}$ This problem involves finding dy/dx for parametric equations using the fundamental formula dy/dx = (dy/dt)/(dx/dt). Given x(t) = t² + 1 and y(t) = √t, we calculate dx/dt = 2t and dy/dt = 1/(2√t). Therefore, dy/dx = (1/(2√t))/(2t) = 1/(4t√t) = 1/( $4t^{3/2}$ ). At t = 4, we get dy/dx = 1/(4· $4^{3/2}$ ) = 1/(4·8) = 1/32. Choice B incorrectly inverts the fraction and gets 32 instead of 1/32. When working with parametric derivatives, maintain the correct order: dy/dt over dx/dt, not the reverse.
A particle moves with $x(t)=\frac{1}{2}t^2$ and $y(t)=\frac{1}{4}t^4$ ; compute $\frac{dy}{dx}$ .
Answer: $\dfrac{t^3}{t}$ This parametric curve requires $dy/dx = (dy/dt)/(dx/dt)$ . For $x(t) = \frac{1}{2}t^2$ , we find $dx/dt = t$ . For $y(t) = \frac{1}{4}t^4$ , we get $dy/dt = t^3$ . Using the parametric derivative formula yields $dy/dx = t^3/t = t^2$ . Choice A shows this result correctly. The standard parametric differentiation method involves finding derivatives of both components, then dividing $dy/dt$ by $dx/dt$ .
For $x(t)=t\cos t$ and $y(t)=t\sin t$ , determine $\frac{dy}{dx}$ in terms of $t$ .
Answer: $\dfrac{\sin t+t\cos t}{\cos t-t\sin t}$ This parametric differentiation uses dy/dx = (dy/dt)/(dx/dt). For x(t) = t cos t, we apply the product rule: dx/dt = cos t - t sin t. For y(t) = t sin t, we get dy/dt = sin t + t cos t. The parametric derivative gives dy/dx = (sin t + t cos t)/(cos t - t sin t). Choice B incorrectly inverts the fraction. The key parametric derivative technique is to differentiate each component separately, then form the ratio with dy/dt in the numerator.

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