University:
Shippensburg University of Pennsylvania
Course:
MAT 211 | Calculus I
Academic year:

2024
Views:

236

Pages:

3
Author:

Herbosambx

Question 1 Rewrite as an exponential equation ln ⁡ 5 = 𝑦 ln5=y Answer Step 1: The relationship between logarithmic and exponential is given by log ⁡ 𝑏 𝑥 = 𝑦 ⟺ 𝑏 𝑦 = 𝑥 log b x=y⟺b y =x Where 𝑏 b is the base. To convert logarithmic function to exponential, find 𝑥 , 𝑦 , 𝑏 x,y,b, and substitute into the above equation. Step 2: Given: ln ⁡ 5 = 𝑦 ln5=y The base of natural logarithm is 𝑒 e. So, 𝑏 = 𝑒 b=e. Here 𝑥 = 5 x=5 and 𝑦 = 𝑦 y=y. So the exponential form is 𝑏 𝑦 = 𝑥 ⟹ 𝑒 𝑦 = 5 b y =x⟹e y =5 Question 2 Convert the following polar equation into a Cartesian equation. Specifically describe the graph of the equation in rectangular coordinates: 𝑟 = 5 sin ⁡ 𝜃 r=5sinθ Answer Step 1: In order to transform from polar to Cartesian. Step 2: Given: 𝑟 = 5 sin ⁡ 𝜃 r=5sinθ Step 3: Formula used: 𝑥 = 𝑟 cos ⁡ 𝜃 , 𝑦 = 𝑟 sin ⁡ 𝜃 , 𝑟 2 = 𝑥 2 + 𝑦 2 x=rcosθ,y=rsinθ,r 2 =x 2 +y 2 Step 4: Simplify as: 𝑟 2 = 5 𝑟 sin ⁡ 𝜃 r 2 =5rsinθ Multiply 𝑟 r as: 𝑥 2 + 𝑦 2 = 5 𝑦 (since, 𝑟 2 = 𝑥 2 + 𝑦 2 and 𝑦 = 𝑟 sin ⁡ 𝜃 ) x 2 +y 2 =5y(since, r 2 =x 2 +y 2 and y=rsinθ) 𝑥 2 + 𝑦 2 − 5 𝑦 = 0 x 2 +y 2 −5y=0 Step 5: Complete the square for 𝑦 y: 𝑥 2 + ( 𝑦 − 5 2 ) 2 = 25 4 x 2 +(y− 2 5 ) 2 = 4 25 Therefore, the equation's Cartesian form is: 𝑥 2 + ( 𝑦 − 5 2 ) 2 = ( 5 2 ) 2 x 2 +(y− 2 5 ) 2 =( 2 5 ) 2 This is the equation of a circle with center ( 0 , 5 2 ) (0, 2 5 ) and radius 5 2 2 5 . Question 3 Is the answer to this system of equations ( 3 , 8 ) (3,8)? 𝑦 = 7 𝑥 + 8 y=7x+8 𝑦 = 𝑥 + 1 y=x+1 True False Answer Step 1: Analysis Given: 𝑦 = 7 𝑥 + 8 ( 1 ) y=7x+8(1) 𝑦 = 𝑥 + 1 ( 2 ) y=x+1(2) Step 2: Simplification From (1) and (2): 7 𝑥 + 8 = 𝑥 + 1 7x+8=x+1 6 𝑥 = − 7 ⟹ 𝑥 = − 7 6 6x=−7⟹x=− 6 7 Substituting 𝑥 = − 7 6 x=− 6 7 into either equation gives a value of 𝑦 y not equal to 8 8. Hence, ( 3 , 8 ) (3,8) is not a solution. Answer: False Question 4 Find parametric equations for the curve with the given properties. The line with slope − 2 −2, passing through ( − 10 , − 20 ) (−10,−20). Answer To find the parametric equation of the line with slope − 2 −2 passing through ( − 10 , − 20 ) (−10,−20): Step 1: Given slope 𝑚 = Δ 𝑦 Δ 𝑥 = − 2 m= Δx Δy =−2. Step 2: The parametric equations of the line are given by: 𝑥 = 𝑥 1 + 𝑡 , 𝑦 = 𝑦 1 + 𝑚 ⋅ 𝑡 x=x 1 +t,y=y 1 +m⋅t Step 3: Substituting the values 𝑥 1 = − 10 x 1 =−10, 𝑦 1 = − 20 y 1 =−20, 𝑚 = − 2 m=−2: 𝑥 = − 10 + 𝑡 , 𝑦 = − 20 − 2 𝑡 x=−10+t,y=−20−2t Hence, the parametric equations are: 𝑥 = − 10 + 𝑡 , 𝑦 = − 20 − 2 𝑡 x=−10+t,y=−20−2t Question 5 Determine the slope of the line represented by the linear equation 5 𝑥 + 6 𝑦 = 40 5x+6y=40: 1.5 1.5 2.6 2.6 − 5 6 − 6 5 4 4 Answer Step 1: Given the equation of the line: 5 𝑥 + 6 𝑦 = 40 5x+6y=40 Step 2: The slope of this line must be determined. Step 3: Rearranging into slope-intercept form 𝑦 = 𝑚 𝑥 + 𝑏 y=mx+b: 6 𝑦 = − 5 𝑥 + 40 ⟹ 𝑦 = − 5 6 𝑥 + 40 6 6y=−5x+40⟹y=− 6 5 x+ 6 40 The slope is: 𝑚 = − 5 6 m=− 6 5 Answer: − 5 6 − 6 5 (Option 3).