4.8E: Exercises for Section 4.8
- Page ID
- 53283
In exercises 1 - 4, each set of parametric equations represents a line. Without eliminating the parameter, find the slope of each line.
1) \(x=3+t,\quad y=1−t\)
2) \( x=8+2t, \quad y=1\)
- Answer:
- \(m=0\)
3) \( x=4−3t, \quad y=−2+6t\)
4) \( x=−5t+7, \quad y=3t−1\)
- Answer:
- \(m= -\frac{3}{5}\)
In exercises 5 - 9, determine the slope of the tangent line, then find the equation of the tangent line at the given value of the parameter.
5) \( x=3\sin t,\quad y=3\cos t, \quad \text{for }t=\frac{π}{4}\)
6) \( x=\cos t, \quad y=8\sin t, \quad \text{for }t=\frac{π}{2}\)
- Answer:
- Slope\(=0; y=8.\)
7) \( x=2t, \quad y=t^3, \quad \text{for } t=−1\)
8) \( x=t+\dfrac{1}{t}, \quad y=t−\dfrac{1}{t}, \quad \text{for }t=1\)
- Answer:
- Slope is undefined; \( x=2\).
9) \( x=\sqrt{t}, \quad y=2t, \quad \text{for }t=4\)
In exercises 10 - 13, find all points on the curve that have the given slope.
10) \( x=4\cos t, \quad y=4\sin t,\) slope = \(0.5\)
- Answer:
- \( t=\arctan(−2); \left(\frac{4\sqrt{5}}{5},\frac{−8\sqrt{5}}{5}\right)\).
11) \( x=2\cos t, \quad y=8\sin t,\) slope= \(−1\)
12) \( x=t+\dfrac{1}{t}, \quad y=t−\dfrac{1}{t},\) slope= \(1\)
- Answer:
- No points possible; undefined expression.
13) \( x=2+\sqrt{t}, \quad y=2−4t,\) slope= \(0\)
In exercises 14 - 16, write the equation of the tangent line in Cartesian coordinates for the given parameter \(t\).
14) \( x=e^{\sqrt{t}}, \quad y=1−\ln t^2, \quad \text{for }t=1\)
- Answer:
- \( y=−(\frac{2}{e})x+3\)
15) \( x=t\ln t, \quad y=\sin^2t, \quad \text{for }t=\frac{π}{4}\)
16) \( x=e^t, \quad y=(t−1)^2,\) at \((1,1)\)
- Answer:
- \( y=2x−7\)
17) For \( x=\sin(2t), \quad y=2\sin t\) where \( 0≤t<2π.\) Find all values of \(t\) at which a horizontal tangent line exists.
18) For \( x=\sin(2t), \quad y=2\sin t\) where \( 0≤t<2π\). Find all values of \(t\) at which a vertical tangent line exists.
- Answer:
- A vertical tangent line exists at \(t = \frac{π}{4},\frac{5π}{4},\frac{3π}{4},\frac{7π}{4}\)
19) Find all points on the curve \( x=4\cos(t), \quad y=4\sin(t)\) that have the slope of \( \frac{1}{2}\).
20) Find \( \dfrac{dy}{dx}\) for \( x=\sin(t), \quad y=\cos(t)\).
- Answer:
- \( \dfrac{dy}{dx}=−\tan(t)\)
21) Find the equation of the tangent line to \( x=\sin(t), \quad y=\cos(t)\) at \( t=\frac{π}{4}\).
22) For the curve \( x=4t, \quad y=3t−2,\) find the slope and concavity of the curve at \( t=3\).
- Answer:
- \( \dfrac{dy}{dx}=\dfrac{3}{4}\) and \( \dfrac{d^2y}{dx^2}=0\), so the curve is neither concave up nor concave down at \( t=3\). Therefore the graph is linear and has a constant slope but no concavity.
23) For the parametric curve whose equation is \( x=4\cos θ, \quad y=4\sin θ\), find the slope and concavity of the curve at \( θ=\frac{π}{4}\).
24) Find the slope and concavity for the curve whose equation is \( x=2+\sec θ, \quad y=1+2\tan θ\) at \( θ=\frac{π}{6}\).
- Answer:
- \( \dfrac{dy}{dx}=4, \quad \dfrac{d^2y}{dx^2}=−6\sqrt{3};\) the curve is concave down at \( θ=\frac{π}{6}\).
25) Find all points on the curve \( x=t+4, \quad y=t^3−3t\) at which there are vertical and horizontal tangents.
26) Find all points on the curve \( x=\sec θ, \quad y=\tan θ\) at which horizontal and vertical tangents exist.
- Answer:
- No horizontal tangents. Vertical tangents at \( (1,0)\) and \((−1,0)\).
In exercises 27 - 29, find \( d^2y/dx^2\).
27) \( x=t^4−1, \quad y=t−t^2\)
28) \( x=\sin(πt), \quad y=\cos(πt)\)
- Answer:
- \( d^2y/dx^2 = −\sec^3(πt)\)
29) \( x=e^{−t}, \quad y=te^{2t}\)
In exercises 30 - 31, find points on the curve at which tangent line is horizontal or vertical.
30) \( x=t(t^2−3), \quad y=3(t^2−3)\)
- Answer:
- Horizontal \( (0,−9)\);
Vertical \( (±2,−6).\)
31) \( x=\dfrac{3t}{1+t^3}, \quad y=\dfrac{3t^2}{1+t^3}\)
In exercises 32 - 34, find \( dy/dx\) at the value of the parameter.
32) \( x=\cos t,y=\sin t, \quad \text{for }t=\frac{3π}{4}\)
- Answer:
- \(dy/dx = 1\)
33) \( x=\sqrt{t}, \quad y=2t+4,t=9\)
34) \( x=4\cos(2πs), \quad y=3\sin(2πs), \quad \text{for }s=−\frac{1}{4}\)
- Answer:
- \(dy/dx = 0\)
In exercises 35 - 36, find \( d^2y/dx^2\) at the given point without eliminating the parameter.
35) \( x=\frac{1}{2}t^2, \quad y=\frac{1}{3}t^3, \quad \text{for }t=2\)
36) \( x=\sqrt{t}, \quad y=2t+4, \quad \text{for }t=1\)
- Answer:
- \(d^2y/dx^2 = 4\)
37) Find intervals for \(t\) on which the curve \( x=3t^2, \quad y=t^3−t\) is concave up as well as concave down.
38) Determine the concavity of the curve \( x=2t+\ln t, \quad y=2t−\ln t\).
- Answer:
- Concave up on \( t>0\).
Contributors and Attributions
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.